1. Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Which one has a greater resistance?

Answer: P=VI=  V2/R For the same V, R is inversely proportional to P. 


Therefore, the bulb 60 W, 220 V has a greater resistance.


2. Question: A torch bulb has a resistance of 1 Ω when cold. It draws a current of 0.2 A from a source of 2 V and glows. Calculate

(i) the resistance of the bulb when glowing and
(ii) explain the reason for the difference in resistance.

Answer:
(i) When the bulb glows:

V = I R ---- Ohm's law R = V/I = 2/.2 =10 Ω


(ii) Resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold.


3. Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of copper = 1.72 x 1 0-8


Answer: L = 1 km = 1000 m

R = 1 mm = 1 x 1 0-3

p = 1.72 x 1 0-8  W m


Area of cross section = p r2  = 3.14 x 1 0-3 x 1 0-3 =  3.14 x 1 0-6


R = pl/A = (1.72 x 1 0-8  x 1000 ) / 3.14 x 1 0-6  = 5.5 W


4. Question: When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1 A is found to flow through it. Calculate: 

(i) the resistance per unit length of the wire                   
(ii) the resistance of 2 m length of this wire 

(iii) the resistance across the ends of the wire if it is doubled on itself.


Answer: (i) V = I R ----- Ohm's law R=V/I=2/1= 2 Ohm

Resistance per unit length: 2/5= 0.4 Ohm/m


(ii) Resistance of 2 m length of the wire = 0.4 x 2=0.8 ohm


(iii) When the wire is doubled on itself:


(a) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.


(b) The length becomes half i.e.L/2 


Resistance of this wire =R' = p (l/2)/(2A) = 1/4(p(L/A)


But p(L/A) = 2 ohm


R' = 1/4 x 2=0.5 Ohm


5. How much work is done in moving 4 C across two point having pd. 10 v


Solution : W = VQ = 10 x 4 = 40J


6. How much energy is given to each coulomb of charge passing through a 9 v battery?


Solution:  Potential difference = Work done = Potential difference × charge

Where, Charge = 1 C and Potential difference = 6 V

Work done = 9×1 = 9 Joule.


7. 100  j of work is done in moving a charge of 5 C from one terminal of battery to another . What is the potential difference of battery?


Solution: V = W/Q =  100j/5C = 20 V


8. If 4 x 10 -3 J of work is done in moving  a particles carrying a charge   of 16 x 10 - 6 C from infinity to point P .What will be the potential at a point?


Solution: the potential at a point is work done to carry unit from one point to another  

                         = (4 x 10 -3 ) /(16 x 10 - 6 C) = 250 V


9. Calculate the current and resistance of a 100 W ,200V electric bulb.

Solution:Power,P = 100W   and     Voltage,V = 200V

Power  P  = VI

So, Current I = P/v = 100/200 = 0.5A

Resistance R = V/I = 200/0.5 = 400W.


10.Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.


Solution:

Voltage ,V = 220V     and  Current ,I = 5A,


Power,P = VI = 220 × 5 = 1100W = 1.1 KW.


11.A lamp can work on a 50 volt mains taking 2 amps.What value of the resistance must be connected in series with it so that it can be operated from 200 volt mains giving the same power.

Solution: Lamp voltage ,V = 50V and  Current ,I = 2 amps.

Resistance of the lamp = V/I   = 50 / 2     = 25 Ω

Resistance connected in series with lamp = r.
Supply voltage = 200 volt. and  Circuit current I = 2 A


Total resistance Rt= V/I  = 200/2    = 100Ω
                         Rt = R + r          =>    100 = 25 + r      =>  r = 75Ω 


12. Calculate the work done in moving a charge of 5 coulombs from a point at a potential of 210 volts to another point at 240 volts


Solution: Potential diffrence  = 210 ­ - 240 = ­30 V 


So, W.= V x  Q = ­30V  x  5C = ­150 Joules


13. How many electrons pass through a lamp in one minute if the current be 220 mA?


Solution:


I  = 220 mA = 0.22 A 


I = Q/T 


0.22 = Q/60 


Q= 0.22 x 60 = 13.2 C 


No of electron carry 1 C charge = 6 x 10 18 


No of electron carry 13.2 C charge = 6 x 10 18 x 13.2 C = 79.2 x 10 18


14.Calculate the current supplied by a cell if the amount of charge passing through the cell in 4 seconds is 12 C ?


Solution:


I = Q/t = 12/4 = 3A



15. A 2 Volt cell is connected to a 1 Ω resistor. How many electrons come out of the negative terminal of the cell in 2 minutes?


Solution: V = IR =>  I = V/R = 2/1 = 2 A


I = Q/t  =>  Q = It = 2 x 2 x 20 = 80 C


No of electron carry 1 C charge = 6 x 10 18 


No of electron carry  80 C  charge = 6 x 10 18 x 80 C = 108  x 10 18   = 1. 08  x 10 20


16. (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?


(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?


Solution

(a) We are given V = 220 V; R = 1200 Ω.
we have the current I = V/R  = 220 V/1200 Ω = 0.18 A.

(b) We are given, V = 220 V, R = 100 Ω.
 we have the current I = V/R =  220 V/100 Ω = 2.2 A.

17. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Solution

We are given, potential difference V = 60 V, current I = 4 A.

According to Ohm’s law, R = V/I = 60/4 =15Ω

When the potential difference is increased to 120 V


the current is given by current = V/R = 120V/15 = 8A

The current through the heater becomes 8 A.

18. A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.


Solution

We are given, R = 4 Ω.


When a wire is doubled on it, its length would become half and area of cross-section would double. T

So,  a wire of length l and area of cross-section A becomes of length l/2 and area of cross section 2A. we have R = ρ(l/A)

R1 = ρ((l/A) / 2A)   where R1 is the new resistance.

Therefore, R1/R = ρ((l/A)/2A) / ρ(l/A) = 1/4

Or, R1 = R/4 = 4Ω/4 = 1Ω

The new resistance of the wire is 1 Ω.


19 . 3.A circuit is made of 0.4 Ω wire,a 150Ω bulb and a 120Ω rheostat connected inseries.Determine the total resistance of the resistance of the circuit. 


Solution: Resistance of the wire = 0.4Ω              

Resistance of bulb = 150Ω
Resistance of rheostat = 120Ω
In series,  Total resistance ,R = 0.4 + 150 +120 = 270.4Ω


20. A current of 0.2 Ampere flows through a conductor of resistance 4.5 Ω. Calculate the potential difference at the ends of the conductor.


Solution:


The potential difference at the ends of the conductor. =  V = IR = 0.2  x  4.5  = 0.9 V

21. A lamp has a resistance of 96 ohms. How much current flows through the lamp when it is connected to 120 volts?



Solution: I = V/R = 120/96 = 1.25 A [V = IR]

The current through the lamp equals 1.25 A.'

22. The manufacturer specifies that a certain lamp will allow 0.8 ampere of current when 120 volts is applied to it. RRWhat is the resistance of the lamp?

Solution: V = IR So, R = V/I = 120/0.8 = 150 W

23. How much voltage is required to cause 1.6 amperes in a device that has 30 ohms of resistance?
Given: V = IR = 1.6 x 30 = 48 V

24. How much power is dissipated when 0.2 ampere of current flows through a 100-ohm resistor?
Ans: P = V I = IR x I = I2 R = 0.2 x 0.2 x 100 = 4 W


25, How much energy is converted by a device that draws 1.5 amperes from a 12-volt battery for 2 hours?

W = Pt, P = V I So, W = VIt = 12 x 1.5 x 2 = 36 W

Important Numerical Questions with Solutions for Chapter Electricity

Ohm's Law
1. A current of 2 A is flowing through a conductor having a resistance of 10 Ω. Calculate the potential difference across the conductor.
Solution: V = IR = 2 × 10 = 20 V

2. A potential difference of 12 V is applied across a conductor having a resistance of 6 Ω. Calculate the current flowing through the conductor.
Solution: I = V/R = 12/6 = 2 A

3. A current of 3 A is flowing through a conductor having a potential difference of 15 V across it. Calculate the resistance of the conductor.
Solution: R = V/I = 15/3 = 5 Ω

4. A current of 4 A flows through a conductor having a resistance of 8 Ω. Calculate the potential difference across the conductor.
Solution: V = IR = 4 × 8 = 32 V

Kirchhoff's Laws
5. In a circuit, there are three resistors of 4 Ω, 6 Ω, and 8 Ω connected in series. Determine the total resistance of the circuit.
Solution: Total resistance (R) = R1 + R2 + R3 = 4 + 6 + 8 = 18 Ω

6. In a circuit, there are three resistors of 12 Ω, 15 Ω, and 18 Ω connected in parallel. Determine the total resistance of the circuit.
Solution: Total resistance (R) = 1/((1/R1) + (1/R2) + (1/R3)) = 1/((1/12) + (1/15) + (1/18)) = 5.45 Ω

7. In a circuit, there are two resistors of 6 Ω and 10 Ω connected in series. A potential difference of 24 V is applied across the circuit. Calculate the current flowing through the circuit.
Solution: Total resistance (R) = R1 + R2 = 6 + 10 = 16 Ω
I = V/R = 24/16 = 1.5 A

8. In a circuit, there are two resistors of 3 Ω and 6 Ω connected in parallel. A potential difference of 12 V is applied across the circuit. Calculate the current flowing through the circuit.
Solution: Total resistance (R) = 1/((1/R1) + (1/R2)) = 1/((1/3) + (1/6)) = 2 Ω
I = V/R = 12/2 = 6 A

Power
9. A current of 4 A flows through a conductor having a resistance of 5 Ω. Calculate the power dissipated by the conductor.
Solution: P = I^2R = 4^2 × 5 = 80 W

10. A potential difference of 10 V is applied across a conductor having a resistance of 2 Ω. Calculate the power dissipated by the conductor.
Solution: P = V^2/R = 10^2/2 = 50 W

Conclusion
These numerical questions are important for class 10 students to understand the concepts of electricity. The solutions provided will help students to solve similar questions easily.