Class 10 Science Chapter 11 Previous Year Questions - Electricity

Ans: (A) When two 6 Ω resistances are connected in parallel and the third resistance of 6 Ω  is connected in series combinations to this, then equivalent resistance will be 9 Ω

OR
(B) Equivalent resistance

Using ohms law :

Ans: (A) (i) Electric power: Rate at which electrical energy is dissipated or consumed / Rate of supplying energy to maintain the flow of current through a circuit.

(ii) (a) Energy consumed = 11 units 

=1100 W

Thus, the power rating of the oven is 2200 W or 2.2 kW.

• Resistivity will not change. 
• because Resistivity does not depend on the dimension of the conductor and only depends on the nature of the material.

Ans: (b) R₁ > R₂ > R₃

Ans:

Ans: (d)
The resistance of a wire is inversely related to its cross-sectional area and directly related to its length. To minimize resistance, you should use a wire with a larger diameter and a shorter length. Therefore, if you choose a diameter of 4D (making the cross-sectional area much larger) and a length of 2L (shorter compared to the original), you will achieve the minimum resistance, making the correct answer (d) 4D and 2L.

Ans: It prevents damage to the appliances and the electrical circuit from overloading and short circuiting.
Here P = 3 kW = 3000 W, V = 220 V, I = ?
P = V I

13·63 A > Rating of fuse 5 A, therefore fuse wire will melt and break the circuit.

Ans: (a) Ohm’s Law: The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same.

(b)

Ans: (a) (i) Bulb A glows  (ii) Bulbs B, C, D and E glow    
(b) P = V × I

(c) (i) Resistance of bulb B,
(alternative formula for calculation
(ii) Total resistance of the series combination of four bulbs = 4 x 275 = 1100 W
OR
(c) 

• Bulb A will keep glowing with same brightness. 
• Other bulbs i.e., B, D and E will stop glowing.
• Reason: As the bulbs B, D and E are connected in series with fused bulb C, so no current flows through them and thus they will not glow. The bulb A remains unaffected as it is connected in parallel combination.

Ans: (c)
To determine which combinations of resistors have an equivalent resistance of 1 ohm, you need to analyze the given arrangements (I, II, III, and IV) based on how resistors are connected, either in series or parallel. Combinations that meet the criteria of having an equivalent resistance of 1 ohm are identified in options I and II. Therefore, the correct answer is (c) I and II.

Ans: (a)
To calculate the heat developed in the electric iron, we can use the formula H = I² Rt, where H is the heat produced, I is the current (5 A), R is the resistance (20 ohms), and t is the time (30 seconds). 
Plugging in the values: H = (5²) × 20 × 30 = 25 × 20 × 30 = 15000J 
So, the heat developed in the iron is 15000 J, making the correct answer (a) 15000 J.

Ans: Wire A will offer more resistance
Justification:
 

• Smaller diameter ⇒ Smaller cross-sectional area ⇒ Higher resistance.
• Wire A (0.2 mm diameter) has higher resistance than wire B (0.3 mm diameter).

Ans: (a) R_s = 4 Ω + 6 Ω + 16 Ω = 26 Ω
(b) (c) (i) Total resistance = 26 Ω + 4 Ω = 30 Ω
Potential difference = V = 6V

OR
(c) (ii) 16 Ω
Justification: According to Ohm’s law when same current flows, the potential difference across a higher resistance is always higher.
Potential difference across 16 Ω = V = IR = 0.2 x 16 = 3.2V
Potential difference across 8 Ω = V = IR_(total) = 0.2 x 4 = 0.8V

Ans: Q = I x t

Ans: (a) (i) 

• Current becomes one-third of its initial value.
• Ohm’s Law

The potential difference across the ends of a conductor is directly proportional to the current flowing through it, provided its temperature remains the same.      
(ii)

Total Voltage = V = 4 x 1·5 V = 6 V
Total resistance, R(s) = R1+ R2 + R3
= 5 Ω + 10 Ω + 15 Ω = 30 Ω

OR
(b) (i) When 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

When the diameter is doubled, d' = 2d A' = 4A

Ans: (d)
To find the total resistance between points X and Y, we need to analyze the circuit configuration.

The given resistances are:
2 Ω, 3 Ω, and 6 Ω.

Looking at the circuit:
The 3 Ω and 6 Ω resistors are connected in parallel.
This parallel combination is then in series with the 2 Ω resistor.
Step 1: Calculate the equivalent resistance of the parallel combination of 3 Ω and 6 Ω.
Using the formula for parallel resistance:

So, R_parallel = 2Ω.
Step 2: Add the series resistance.
Now, the 2 Ω (from the parallel combination) is in series with the 2 Ω resistor at the start:
R_total = 2 + 2 = 4Ω
Therefore, the total resistance between points X and Y is 4 Ω.
The correct answer is (b) 4 Ω.

Ans: 

Equivalent resistance in the circuit: 
R = 2 + 6 + 16 
= 24 Ω 
The voltage supplied by the battery: 
V = 4 × 1.5 = 6 V 
(A) Using Ohm’s law we can calculate ammeter’s reading I. 
I = V / R = 6 / 24 = = 0.25 A 
(B) Voltmeter’s reading is: 
V' = IR = 0.25 ×16 = 4 V

Ans: (b)
The correct expressions that relate charge (Q), current (I), time (t), voltage (V), and work done (W) are as follows: 
For the relationship between charge, current, and time: Q = I × t (the total charge is the product of current and time). 
For the relationship between work, voltage, and charge: W = V × Q (the work done is the product of voltage and charge). 
Thus, the correct answer is (b) (i) Q = I × t and (ii) W = V × Q.

Ans: (i) It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.

(ii) To measure the entire current passing through the circuit, the ammeter should have low resistance.
(iii) R₅ = R₁ + R₂ = Maximum resistance

So, R_A > R_B

Thus, A shows series combination and B shows parallel combination.

Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit.
Power = Work/Time = Energy/Time
The SI unit of electric power is watt (W).
1W = 1 volt x 1 ampere = 1
VA 1 unit = 1 kWh = 3.6 x 10⁶ J
(b) 2 bulbs: P = 50 W, t = 6 hr daily
1 geyser: P = 1 kW, t = 1hr
E=P x t used by every 2 bulbs + energy used by geyser
= 30 days, Rs. = 8/kWh
Now, total energy consumed in one day
= 2 × 50 × 6 + 1 x 1000 1600 Wh
Total energy consumed in 30 days
= 30 x 1600 Wh = 48 kWh
Bill = 8 x 48 = Rs. 384.

Ans: (d)
Given:
Four identical resistors, each with a resistance of 8Ω.
Calculate the total resistance in series ($\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}R\_S$R_S):When resistors are connected in series, the total resistance is the sum of individual resistances:
R_S = 8 + 8 + 8 + 8 = 4 × 8 = 32Ω
Calculate the total resistance in parallel (R_P):When resistors are connected in parallel, the total resistance is given by:

So, R_P = 2Ω
Calculate the ratio R_S / R_P

Therefore, the correct answer is (d) 16.

Ans: (c)
To determine the least count of the voltmeter and the reading shown by the needle, let's analyze the diagram.

Least Count Calculation:

• The scale on the voltmeter ranges from $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}$0 to $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}3$3 volts.
• There are 15 divisions between $0$0 and $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}1.5$1.5 volts, and similarly 15 divisions between 1.5 and $3$3 volts.
• Therefore, each division represents: 1.5V / 15 = 0.1V
• So, the least count of the voltmeter is 0.1 V.

Reading of the Voltmeter:

• The needle is positioned at 1.8 V on the scale.

Based on this analysis, the correct answer is:

• Least count: 0.1 V
• Reading shown by the voltmeter: 1.8 V

Thus, the correct option is (c) 0.15 V and 1.8 V.

Ans: (A) Power = VI = V² / R, 
where V is voltage, I is current, R is resistance. 
First case, 
Power = 880 W 
Voltage = 220 V 
⇒ 880 = 220 × I 
⇒ I  = 4 A 
⇒ 880 = 220² / R 
⇒ R = 55 Ω 
Second Case, 
Power = 330 W
= 220 × I 
⇒ I = 3 / 2 A 
⇒ 330 = 220² / R 
⇒ R = 440 / 3 Ω
(B) When a conductor provides resistance to current flow, the work done by the current in overcoming this resistance is converted into heat energy. This is known as the current heating effect.
(C) The amount of work done W in carrying a charge Q through a wire of resistance R in time t is given by: 
W = QV Since Q = I × t 
Therefore. 
W = V × I × t 
where V is the potential difference across the wire. 
Since by Ohm’s law, 
V = IR
Therefore, W = I²Rt 
The electric energy dissipated or consumed is directly proportional to the square of the current I, directly proportional to the resistance R and to the time t during which current flows. 
H = I²Rt

Ans: (a) Ohm's law states that the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided physical conditions like temperature etc., are kept unchanged.
Mathematically, V ∝ I
or V/I = constant or V/I = R ⇒ V = IR
where, R is called the resistance of the conductor. SI unit of resistance is ohm and is denoted by Ω. It is a constant of proportionality and its value depends upon the size, nature of material and temperature. (b) If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is said to be 1Ω.
We know that, R = V/l
Therefore, 1ohm =
(c) Given, potential difference V = 2 V
Current, 1 = 0.5 A
Using Ohm's law, V = IR

Ans: First, calculate the resistance of 2 series resistors inside the loop, i.e., = R₁ +R₂
 =10Ω + 10Ω
= 20Ω
To calculate the equivalent resistance in the given electric circuit, let us find the parallel resistance. For that we use

Now, again applying the series formula to add the resistors together =10Ω + 10Ω + 20Ω = 40Ω
So the total resistance of the given 40Ω

Ans:  (i)
We know that resistance R₁, R₂ and R₃ are in series.
So, R_AB = R₁ + R₂ + R₃

So,

or

(ii) The slope of V-I graph gives the resistance. The greater the slope greater will be the resistance. We know, R_s > R_p 
∴ Graph (I) is correct

Ans: The equivalent resistance in the arm A = 5Ω + 15Ω + 20Ω =40 Ω
(b) The equivalent resistance of the parallel combination of the arms B and C= ((1/10Ω + 20Ω + 30Ω) + (1/5Ω + 10Ω + 15Ω))^-1 =20Ω
(c) (i) Current flow flowing through the ammeter = 6/60 = 0.1 A
OR
(ii) Current that flows in the ammeter when the arm B is withdrawn
from the circuit = 6/100 = 0.06A

Ans: (i) According to Joule's law of heating, when a current (I) is passed through a conductor of resistance (R) for a certain time (t), the conductor gets heated up and the amount of energy released is given by

Ans: (a) Here, the length of the wire, l = 2m
Area of cross-section, A = 1.55 × 10^-6 m² 
Resistivity of metal, ρ = 2.8 × 10^-8 Ω m

∴

Ans: Power of electric heater, P = 1100 W
Operating voltage, V = 220 V
(i) Its resistance, R= V²/P
R = 220 x 220/1100 Ω : R = 44 Ω
operating voltage, V = 220 V
So, current I = V/R = 220/44= 5A

Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit.
Power = Work/Time = Energy/Time
The SI unit of electric power is watt (W)
1 W =1 volt × 1 ampere = 1 V A
(b) (i) Electrical energy is the product of power and time.
E = P x t
= 2 kW × 2h = 4kW h
(ii) Electric energy consumed in joules
1 kWh =3.6× 10⁶ J
4 kW h = 3.6 × 10⁶ x 4 = 14.4 x 10⁶ J.

Ans: (i) The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power.
SI unit of electric power is watt (W).
(ii) Given, P₁ = 100 W, V₁ = 220 V
P₂ = 60 W,V₂ = 220 V
P = VI

∴ Total current = l₁ + l₂ = 0.45 + 0.27 = 0.72 A

Ans: Electric Power: The rate at which electrical energy is consumed or dissipated is called electrical power.

P = VI,
[∵ V = IR]
P = (IR) I = I² R

Ans: Current drawn by 1st lamp rated 100W at 220V

and current drawn by 2nd lamp rated 60W at 220V

In parallel arrangement the total current

Ans: (a)
Sol: The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,

Concept Applied Combination of resistances in series have maximum value and combination of resistances in parallel have minimum value.

Ans: (c)
Sol: A group of resistors can produce maximum resistance when they all are connected in series.

Ans: (a)
Sol: The maximum resistance can be produced from a group of resistors by connecting them in series.
Thus,

Ans: V-I graph: The nichrome cable's V-I graph exhibits a straight line. It signifies that the resistance of the wire stays unchanged while the current supply has been varied. That is an ohmic conductor because it follows the ohm's law.
Mathematically, ohm's law can be represented as:
V = IR
Here,
R is the resistance.
V is the voltage.
I is the current.
The resistance of a wire will be:
⇒ R = V/I
= 0.4/0.1
= 4Ω
Thus, the circuit diagram for the given case is,

Therefore, because nichrome wire has a constant resistance of 4 but also obeys Ohm's law, it is classified as an ohmic conductor.

Ans: (a) As per mathematical expression for Joule.s law of heating
Heat produced H = VIt = I2 Rt = V²/R t
When a voltage V is applied across a resistance R for time t so that a current I flows through it.
(b) Here charge Q = 96000 C, time t = 2 h and potential difference V = 40 V
Heat generated H=VIt = VQ
⇒ H = 40 x 96000 = 3840000 J = 3.84 x 10⁶ J

Ans: 
Sol: Current through bulb:
The same current will flow through both bulbs in series.
When the same current flows through all of the circuit's components, the circuit is said to be connected in series. The current has only one path in such circuits.
As an example of a series circuit, consider the household decorative string lights. This is nothing more than a string of tiny bulbs connected in series.
If one of the bulbs in a series burns out, none of the others will light up.
Hence, Option(D) is correct.

Ans: (a) If a current of 1 ampere flows through it when the potential difference across it is 1 volt
(b) The electric power is the electric work done per unit time [P=W/T]. the relation between electric powe , potential difference and resistance is P = V²/R
(c) R= V/I
= 220/5 = 44 ohm
R= 44 = 132/n
hence, 132/44 = 3 Resistors 

Ans: (a) The Joule's law of heating implies that heat produced in a resistor is
(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor. i.e., H = I² Rt
(b) Given, charge q = 96000C time t = 2h = 7200s and potential difference V = 40V
We know,

Ans: (a)

For a conductor, resistance R is given by:

where:

• ρ is the resistivity of the material (constant for a given material),
• l is the length of the conductor,
• A is the area of cross-section.

Given:

1. A conductor with length $l$l, area A, and resistance $R$R.
2. Another conductor of the same material with length 2.5l and resistance 0.5R.

Let the area of cross-section of the second conductor be A′.

Since the two conductors are made of the same material, their resistivity ρ is the same. We can set up the resistance formulas for each conductor:
For the first conductor:

For the second conductor:

Now, substitute R = ρ⋅l / A into the equation for the second conductor:

Canceling ρ and l from both sides:

Rearranging to solve for A′:
A′ = 2.5 × 0.5A = 5A

Therefore, the area of cross-section $A\prime A\text{'}$A′ of the second conductor is 5A.

Ans: (a)

The heating effect $H$H in a resistor is given by Joule's law of heating:

H = I²Rt

where:

• $I$I is the current,
• $R$R is the resistance,
• $t$t is the time.

If the resistance $R$R is reduced to half of its initial value, and assuming the voltage $V$V across the resistor remains unchanged, the current I through the resistor will increase.

Using Ohm’s law:  I = V/R
When $R$R is reduced to $\backslash frac\{R\}\{2\}$R/2, the new current $I\text{'}$I′ becomes:

Now, substituting into the heating formula:

Thus, the heating effect becomes twice the initial heating effect.
Therefore, the correct answer is (a) two times.

Ans: (a)
Assertion (A): Alloys are commonly used in electrical heating devices like electric irons and heaters. This is true because alloys have certain properties that make them suitable for these applications, such as high resistivity, which allows them to produce heat effectively.
Reason (R): "Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals." This statement is also generally true. Alloys like nichrome have a higher resistivity, which is beneficial for heating applications, and they also tend to have controlled melting points, which can sometimes be lower than certain individual metals but still appropriate for their purpose in heating devices.\
Therefore, both the assertion and reason are correct, and the reason correctly explains why alloys are used in heating devices.
So, the correct answer is (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).

Ans: (b)
Assertion (A): "At high temperatures, metal wires have a greater chance of short circuiting." This statement is true. At high temperatures, metal wires can overheat, which may lead to insulation breakdown, increasing the risk of short circuits.
Reason (R): "Both resistance and resistivity of a material vary with temperature." This statement is also true. The resistance and resistivity of metals generally increase with temperature due to increased atomic vibrations, which impede the flow of electrons.
However, the reason is not the direct cause of the assertion. The increased risk of short circuiting at high temperatures is mainly due to the possibility of insulation breakdown and not solely because of changes in resistance or resistivity.
Therefore, the correct answer is (b) Both (A) and (R) are true, and (R) is not the correct explanation of (A).

Ans: (A) 
V = 200 V
P = 100 W
R = ?
P = VI
V = IR

(B) P = 100 W (given 3 such bulbs)
= 100 / 1000  = 0.1 kW
No. of bulbs = 3
time, t = 10 hours
Number of days in the month of November
= 30
E = P × t
Total energy consumed by 3 bulbs in 30 days
= 3 × 0.1 × 10 × 30
= 90 kWh
Hence, the energy consumed by 3 such bulbs for the month of November will be 90 kWh.
(C) Total cost = ? 
Rate of per unit = ₹ 6.50 per unit 
1 kWh = 1 unit 
90 kWh = 90 units 
Tost cost = 90 × 6.50 = ₹ 585.00 
Hence, the total of operating 3 such bulbs will  ₹ 585.

Ans: (c)
Sol: Resistance is given by the slope of V-I graph.
Here the graph is I-V graph.
So, resistance is given by the inverse of the slope.
From the graph, the order of the slope is slope of R₁ > slope of R₂ > slope of R₃
Therefore order of resistance is R₃ > R₂ > R₁.
Hence, option (c) is correct.

Ans: Potential difference between two points of an electric circuit is said to be 1 volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.

Ans: Resistance of a conductor is the measure of opposition offered by it for the flow of electric charge through it. SI unit of resistance is ‘ohm’ (Ω).

Ans: According to Ohm’s law, temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference across its ends, i.e.,
V ∝ I or V = IR
Here, constant R is known as the resistance of given conductor.
Alternative form of Ohm’s Law:
It states that, the current density(J) through a conductor is proportional to the Electric Field(E) across it.
J = σE
Where σ is the conductivity of the conductor.

Ans:

Here, R₁ = 10Ω, R₂ = 20Ω
(a) Current through R₁, l₁ = V/R₁ = 12/10 =1.2A
Current through R₂. I₂ = V/R₂ = 12/20 = 0.6 A
Current through R₃, I₂ = V/R₃ = 12/30 = 0.4A
(b) Total circuit resistance

Ans: Resistance is defined as the opposition to the flow of electrical current through a conductor. The resistance of an electric circuit can be measured numerically. Conductivity and resistivity are inversely proportional. The more conductive, the less resistance it will have.
Resistance = Potential difference/ Current
Factors on which the conductor depends

• The temperature of the conductor 
• The cross-sectional area of the conductor 
• Length of the conductor 
• Nature of the material of the conductor

Electrical resistance is directly proportional to the length (L) of the conductor and inversely proportional to the cross-sectional area (A). It is given by the following relation.
R = ρl/A
where ρ is the resistivity of the material (measured in Ωm, ohm meter)
Resistivity is a qualitative measurement of a material’s ability to resist flowing electric current. Obviously, insulators will have a higher value of resistivity than of conductors.

Ans: (a) Here power of bulb P = 40 W and voltage V = 220 V
∴ Current drawn by the bulb
and resistance of the bulb R =
(b) On taking another bulb of power P' = 25 W and voltage V = 220 V, there is a change in the value of current and resistance because their values depend on the power of the bulb.
New current
and new resistance

Ans: Amount of work done in carrying a charge Q through a potential difference V is
W = QV.
But Q = It  ∴ W = VIt
As power is defined as the rate of doing work, hence

If R is the value of resistance of the conductor, then V - RI and hence
P = VI = (RI)I = I²R
Again
Thus, in general, we can say that electric power is given by

Ans: When a 2 Ω resistor is joined to a 6 V battery in series with 1 Ω and 2 Ω resistors, total resistance of the combination R_s = 2 + 1 + 2 = 5Ω
∴ Current in the circuit I₁ =
∴ Power used in the 2Ω resistor P₁ =
(ii) When 2 Ω resistor is joined to a 4 V battery in parallel with 12 Ω and 2 Ω resistors, current flowing in 2 Ω resistor is independent of the other resistors.
∴ Current flowing through 2Ω resistor
∴ Power used in the 2Ω resistor
∴

Ans:

In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
According to Ohm ’s law, we have

If the total potential difference between A and B is V, then


Let the equivalent resistance be R, then
V = IR
and hence
⇒

Ans: (a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as:

where ρ is a constant, which is known as the electrical resistivity of the material of conductor.
Thus, resistivity ρ = RA/l
∴ SI unit of resistivity ρ shall be  = Ω-m
(b) Here l = 5 m, R = 100 Ω and A = 3 x 10^-7 m²
∴ Resistivity of the material of wire ρ = RA/l =

Ans: (a)

In parallel combination of three resistances R₁ R₂ and R₃, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I₁ I₂ and I₃. Thus,
I = I₁ + I₂ + I₃.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) Here R₁ = R₂ = 12 Ω and V = 6 V
Net resistance R o f the parallel grouping is :
∴ Current drawn by the circuit from the battery I =

Ans: (i) In the circuit resistor of R₁ = 12 Ω is connected in series with parallel combination of two resistors R₂ and R₃ of 24 Ω each.
The effective resistance of parallel combination of R₂ and R₃ is given as :

∴ Net resistance of the circuit R = R₁ + R₂₃ = 12 + 12 = 24 Ω
∴ Current through 12 Ω resistor = Circuit current
(ii) Both ammeters A₁ and A₂ give same reading of 0.25 A and there is no difference in their readings.

Ans:

(i) The schematic diagram is given in Fig. 12.25.
(ii) Here total voltage V = 5 x 2 = 10 V
and total resistance R = R₁ + R₂ + R₃
= 5 + 10 + 15 = 30 Ω
∴ Current passing through the circuit when the key is closed
(iii) Potential difference across resistor R3 of 15 Ω

Ans: Here voltage of battery V = 6 V, resistance of electric lamp = R₁ = 20 Ω  and resistance of conductor R₂ = 4 Ω.
(a) Since R₁ and R₂ are connected in series, the total resistance of the circuit
R = R₁ + R₂ = 20 + 4 = 24 Ω
(b) The current through the circuit I = V/R = 6/24 = 0.25 A
(c) (i) Potential difference across the electric lamp V1 = IR₁ = 0.25 x 20 = 5 V
(ii) Potential difference across the conductor V₂ = IR₂ = 0.25 x 4 = 1 V
(d) Power of the lamp P = I²R₁ = (0.25)²  x 20 = 1.25 W

Ans: 

(a) The arrangement is shown in circuit diagram of Fig.
In parallel combination of three resistances R₁ R₂ and R₃, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I₁ I₂ and I₃. Thus,
I = I₁ + I₂ + I₃.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) In the network resistors of 20 Ω and 20 Ω are joined in parallel and make a resistance R₁, where
This combined resistance R₁ = 10 Ω is joined in series with given 10 Ω resistance. Hence equivalent resistance of the network will be R = 10 + 10 = 20 Ω.

Ans: (a) (i) In series arrangement, equivalent resistance R_s = R₁ + R₂ + R₃
(ii) In parallel arrangement, equivalent resistance R_p is given as:

(b) Here = R₁ = R₂ = 12 Ω and V = 3V.
For minimum resistance, two resistors must be connected in parallel so that

Hence power

For maximum resistance, two resistors must be connected in series so that
R_s = R₁ + R₂ = 12 + 12 = 24 Ω
So, the power

⇒

Ans: Series combination o f resistors : We take three resistors R₁ R₂ and R₃ and join them in series between the points X and Y in an electric circuit as shown in Fig. 12.42.

(a) Plug the key and note the ammeter reading.
Then change the position of ammeter to anywhere in between the resistors and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flows through each resistor.

(b) 

• Insert a voltmeter across the ends X and Y of the series combination of resistors. 
• Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.
• Take out plug from key K and disconnect the voltmeter. Now insert the voltmeter across the ends of first resistor R₁ as shown in Fig. 12.43. 
• Plug the key and note the voltmeter reading V₁. Similarly, measure the potential difference across the other two resistors R₂ and R₃ separately. 
• Let these potential differences be V₂ and V₃, respectively. Experimentally we find that
  V = V₁ + V₂  + V₃
• It shows that in series arrangement of resistors total potential difference is equal to the sum of potential differences across individual resistors.

Ans: (a)
The unit of electric power can be expressed in terms of volt-ampere (V·A), which represents the power calculated as the product of voltage (in volts) and current (in amperes). This is particularly common in electrical engineering, where 1 watt = 1 volt × 1 ampere.
Here's an explanation of the other options:
(b) kilowatt-hour: This is a unit of energy, not power. It represents the amount of energy used over time (1 kilowatt of power used for 1 hour).
(c) watt-second: This is also a unit of energy, as it represents the power used over a second (1 watt of power used for 1 second).
(d) joule-second: This is a unit related to angular momentum or action in physics, not specifically for electric power.
Therefore, the correct answer is (a) volt-ampere.

Ans: (b)
Using Ohm’s law:
V = I × R
where:
V = 4V (voltage),
I = 100mA = 0.1A (current).
We can rearrange the formula to solve for $RR$R:

Therefore, the resistance of the resistor is 40 Ω.

Ans: (A) Least count of ammeter = 0.5/10 = 0.05 A 
Thus, value corresponding to 12 divisions = 0.05 ×12 = 0.6 A 
(B) An ammeter is connected in series and a voltmeter is connected in parallel in an electric circuit.

Ans: Here resistances R₁ - R₂ = R₃ = 9Ω
(i) To obtain an equivalent resistance R_eq = 13.5Ω, we connect one resistor R₁ in series to the parallel com bination o f R₂ and R₃ as shown in figure (i). Then


(ii) To obtain equivalent resistance R_eq = 6 Ω, we connect resistor R₁ in parallel to the series combination of R₂ and R₃ as shown in figure (ii). Then

Ans: A s per Joule’s law the heat produced in a resistor is (i) directly proportional to square of current flowing through it, (ii) directly proportional to resistance, and (iii) directly proportional to time. Mathematically,
Heat H = I²Rt

Ans: 

• The electrons in the metal are loosely bound while the electrons in the glass are tightly bound together. 
• Glass has one of the lowest possible heat conduction a solid. A good electrical conductivity is the same as a small electrical resistance. 
• Silver is the best conductor because its electrons are freer to move than those of the other elements.  
• When electricity is applied to the metal, ions start to migrate from one end to the other end of the metal, but it is not possible for electrons to travel freely in glass in which electrons are tightly bound. 
• Glass is a bad conductor of electricity as it has high resistivity and has no free electrons.

Ans: Nichrome is an alloy of high resistivity and high melting point and does not oxidise easily.

Ans:

In parallel combination of three resistances R₁ R₂ and R₃, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I₁ I₂ and I₃. Thus,
I = I₁ + I₂ + I₃.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and

Ans: Rubber is an electrical insulator. Hence electrician can work safely while working on an electric circuit without a risk of getting any electric shock.

Ans: For maximum current flow, the net resistance of circuit must be least possible. Hence resistors R₁ and R₂ should be connected in parallel.

Ans: Potential difference V = R x I = 18 x 5 = 90 V.

Ans: We know that the power can be stated as the amount of electric energy consumed per unit time.

On solving For Energy, we get
Energy, E = P/T
Here; P = 60W
t = 1s
E = 60W x 1sec
∴ E = 60J
Thus, the energy in joules is 60J.

Ans: (a) The resistance of a conductor is a property of the conductor, which affects the flow of current through it on maintaining a potential difference across its ends.
Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm, if a potential difference of 1 V is to be applied across its ends for maintaining flow of 1 A current.
(b) If given resistance wire is replaced by another thicker wire o f same material and same length then cross-section area of wire is increased and consequently its resistance decreases  So the current flowing in the circuit increases.

Ans: (a) Electric bulbs are generally filled with some inert gas like nitrogen or argon. This enables to prolong the life of the filament of electric bulb.
(b) Here radius of wire r = 0.01 cm = 0.01 x 10^-2 m, resistance R = 10 Ω and resistivity p = 50 x 10^-8 Ω-m.

⇒

Ans: (a) Electric power is defined as the rate of supplying electrical energy for maintaining current flow through a circuit.
Electric power P = VI = I²R = V²/R.
(b) We know that slope of V-I graph for a given wire gives its resistance and in given figure slope of graph is more for wire A. It means that R_A > R_B.
In series arrangement same current I flows through both the resistance.
As heat produced per unit time is given by I²R, hence it is obvious that more heat will be produced per unit time in wire A.

Ans: 
(a) The arrangement is shown in circuit diagram of Fig. 12.39.
In parallel combination of three resistances R₁ R₂ and R₃, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I₁ I₂ and I₃. Thus,
I = I₁ + I₂ + I₃.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
If R is the equivalent resistance then,
I = V/R
∴
and
(b) Here series combination of R₁ and R₂ is joined to R₃ in parallel arrangement. Hence the net resistance R between points A and B is given as

⇒ r = 4 Ω

Ans: (a)

• In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
• According to Ohm ’s law, we have
  
• If the total potential difference between A and B is V, then
  
  
• Let the equivalent resistance be R, then
  V = IR
  and hence
  ⇒
  (b) Here = R₁ = 2 Ω R₂ = 3 Ω and R₃ = 6 Ω
  The equivalent resistance R for parallel combination of resistors will be
  

Ans: (a) A rheostat, (b) A closed plug key.

Ans: Time rate of flow of charge through any cross-section of a conductor is called electric current.

Ans: An ampere (A), which is defined as the rate of flow of 1 coulomb of charge per second.

Ans: In an electric circuit the direction of flow of electrons is opposite to the direction of flow of current. The direction of flow of current is considered to be the direction of flow of positive charge known as conventional current. The current flows from higher potential to lower potential in an electric circuit .

Ans: Amount of work done to bring 1 C charge from point B to point A in the electric field is 1 joule.

Ans: A kilowatt hour (kW h) is the commercial unit of electrical energy. It is the energy consumed when 1 kW (1000 W) power is used for 1 hour.

Ans: 3.6 x 10⁶ J = 1 kW h.
1 kWh = 1000 × joule/sec × 60 × 60 sec
⇒ 1 kWh = 1000 × 60 × 60 joule/sec × sec
⇒ 1 kWh = 1000 × 60 × 60 joule
⇒ 1 kWh = 36, 00, 000 joule

Ans: 

• A continuous and closed path of an electric current is called an electric circuit.
• A labelled, schematic diagram of an electric circuit showing a cell E, a resistor R, an ammeter A, a voltmeter V and a closed switch S is shown here.
• An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
• The circuit is said to be a closed circuit when the switch is in 'on' mode (or key is plugged) and a current flows in the circuit.
  

Ans: (a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case
electric current I =
As electrons are flowing from east to west, the direction of electric current is from west to east.
(b) Here current I = 1 A, time t = 1 s and charge on each electron e = 1.6 x 10^-19 C.
Hence, number of electrons flowing n =

Ans: (a) The resistance of a cylindrical conductor i.e., a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depends on the nature of material of wire. Mathematically,

Here, ρ is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposite faces.
(b) The resistivity of the conductor remains unchanged.

Ans: (a) Resistance R of wire A is more than that of B (R_A > R_B) because slope of V-I graph for A is more.
The resistance of a wire is inversely proportional to its cross-section area.
Hence area of cross-section of wire B, of smaller resistance, must be more. Thus, wire B is thicker.
(b) (i) The new resistance of wire
Thus, resistance increases to four times its original value.
(ii) The resistivity remains unchanged because it does not depend on the dimensions of a conductor of given metarial.

Ans: (i) Here V = 3 V, R₁ = 10 Ω and R 2 = 20 Ω. Since R₁ and R₂ are connected in series, the effective resistance of the circuit
R = R₁ + R₂ = 10 + 20 - 30 Ω
∴ Current flowing in the circuit
(ii) The potential difference across R 1 = 10 Ω resistor is V₁ = IR₁ = 0.1 x 10 = 1.0 V.

Ans: In the electric circuit shown the series combination of R₁ and R₂ is joined in parallel to the resistance R₃. Hence equivalent resistance R of the circuit is

⇒ R = 2 Ω
∴ Current drawn from the battery I=

Ans: Here V = 6 V and R = 5 Ω
(i) The current flowing through the resistor I =
(ii) Energy dissipated as heat in time t = 10 s is
∴ H = I²Rt = (1.2)² x 5 x 10 = 72 J

Ans: Here charge transferred Q = 90000 C, potential difference b etw een the terminals o f battery V = 40 V and time t = 1 h = 3600 s.
Current =
Amount of heat generated H = Vlt = 40 x 25 x 3600 = 3600000 J = 3.6 x 10⁶ J
and power expended

Ans: The current drawn by a 40 W, 220 V electric lamp
As the electric line is o f rating 220 V, 5 A, hence we can connect n lamps in parallel where

Thus, we can safely connect 27 lamps of 40 W, 220 V rating to a 220 V, 5 A line.

Ans: 

• Electric current is defined as the rate of flow of electric charge through a cross- section of a conductor. 
• If Q charge passes through a section of a conductor in time t, then-current I = Q/t.
• SI unit of electric current is an ampere (A). Current is said to be one ampere, if rate of flow of charge through a cross-section of conductor be 1 coulomb per second.
• Direction of conventional current is taken as the direction of flow of positive charge or opposite to the direction of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B.
• Here current I = 1 A, time t = 1 s and charge on electron e = 1.6 x 10^-19 C. 
• Let n electrons flow through a section of conductor so that charge passing through the section is Q= ne.
  ∴

Ans: When a current flows through a conducting wire (resistance wire), heat is developed, and the temperature of the wire rises. It is known as the heating effect of electric current.
If V is the potential difference maintained across the ends of a wire then, by definition, the amount of work done for flow of 1 C charge through the wire is V.
∴ Work done for flow of Q charge
W = VQ = VIt  [∵ Q = It]
where I is the current flowing in time t.
As V = IR, hence
W= VIt = (IR)It = I²Rt
This work done (i.e., electrical energy dissipated) is converted into heat. Hence, the amount of heat produced, Q = I²Rt J
This is known as Joule’s law of heating.
Incandescent lamps, electric irons, electric stoves, toasters, geysers, electric room heaters, etc., are appliances based on the heating effect of electric current.

Ans: (b)
To achieve the minimum resistance, all resistors should be connected in parallel because the equivalent resistance of resistors in parallel is always less than any individual resistor.
Given:
Each resistor has a resistance of 1/5Ω.
There are five resistors.

When resistors are connected in parallel, the equivalent resistance R_eq is given by:

Since each resistor has the same resistance $R\; =\; \backslash frac\{1\}\{5\}\; \backslash ,\; \backslash Omega$R = 1/5 Ω:

Thus:

$R\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{1\}\{25\}\; \backslash ,\; \backslash Omega$R_eq = 1 / 25 Ω
Therefore, the minimum resistance that can be made using five resistors, each of 1/5Ω, is 1/25Ω.

Ans: An ideal ammeter is one which has zero resistance. But that is not possible. Therefore, the resistance of an ammeter should be as close to zero as possible. If it is non-zero and substantial, it will affect the current flowing through the circuit. This is because an ammeter is connected in series in the circuit for the measurement of electric current.