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All questions of Current Electricity for NEET Exam

A wire of resistance 12 Ωm-1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite point, A and B as shown in the figure, is [2009]
  • a)
    3Ω
  • b)
    6π Ω
  • c)
    6Ω
  • d)
    0. 6π Ω
Correct answer is option 'D'. Can you explain this answer?

Ram Mohith answered
The two halves (or semicircles) are in parallel combination.

Length of each part is πr = π(0.1) m
Resistance of each part = 12(0.1π) = 1.2π

We know that when two equal resistance are kept in parallel the equivalent resistance will be half of any one resistance. So, the equivalent resistance between A and B is 0.6π
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In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is
  • a)
    –1V
  • b)
    + 2V
  • c)
    –2V
  • d)
    + 1V
Correct answer is option 'D'. Can you explain this answer?

Riya Banerjee answered
Current from D to C = 1A
∴ VD – VC = 2 × 1 = 2V
VA = 0     ∴ VC = 1V, ∴ VD – VC = 2
⇒VD – 1 = 2     ∴ VD = 3V
∴ VD – VB = 2 ∴ 3 – VB = 2 ∴ VB = 1V

The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again
  • a)
    All the four resistance should be changed
  • b)
    Both the resistance R1 and R4 should be changed
  • c)
    No resistance needs to be changed
  • d)
    Resistance R4 should be changed only
Correct answer is option 'C'. Can you explain this answer?

Neha Sharma answered
The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
 
Explanation:

  • Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.

  • Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.

  • Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.

  • Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.

  • Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.


    •  

Potentiometer measures the potential difference more accurately than a voltmeter, because
  • a)
    It draws a heavy current from external circuit.
  • b)
    It does not draw current from external circuit.
  • c)
    it has a wire of low resistance.
  • d)
    it has a wire of high resistance
Correct answer is option 'B'. Can you explain this answer?

Amar Pillai answered
Explanation:Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.

A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is
  • a)
    nG
  • b)
    (n−1)G
  • c)
    G/n
  • d)
    G/(n−1)
Correct answer is option 'B'. Can you explain this answer?

Avantika Mehta answered
Understanding Voltmeter Conversion
To convert a voltmeter of range V volts into a voltmeter of range nV volts, we need to consider the resistance of the voltmeter and the additional resistance required in series.
Principle of Operation
- A voltmeter measures the potential difference across its terminals.
- When converting to a higher range, we add a series resistance to ensure that the voltmeter can handle the increased voltage without damaging its internal components.
Given Parameters
- Resistance of the voltmeter = G ohm
- Original range of the voltmeter = V volts
- New desired range = nV volts
Calculating Series Resistance
1. Voltage Division: When a voltmeter with resistance G is connected in series with another resistance R, the voltage drop across the voltmeter is a fraction of the total voltage.
2. Using Voltage Ratios: The voltage across the voltmeter can be given by the ratio:
- V / (V + R) = G / (G + R)
3. Setting Up the Equation: For the new range of nV:
- nV / (nV + R) = G / (G + R)
4. Solving for R: Rearranging gives us:
- R = (n-1)G
This means the resistance required in series to achieve the desired range of nV volts is (n - 1)G ohms.
Conclusion
Thus, the correct answer for the resistance used in series to convert the voltmeter into one of range nV volts is option B: (n - 1)G. This ensures proper functioning and safety of the voltmeter under higher voltage conditions.

A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance
  • a)
    of each of them decreases
  • b)
    of each of them increases
  • c)
    copper increases and that of germanium decreases
  • d)
    copper decreases and that of germanium increases
Correct answer is option 'D'. Can you explain this answer?

Pranjal Pillai answered
Explanation:Copper is a conductor and we know that for conductors, resistance is directly proprtional to temperature. Therefore on decreasing temperature resistance also decreases.Whereas, germanium is a semiconductor and for semiconductors, resistance is inversely proportional to temperature. So on decreasing temperature resistance increases.

In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X<Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y ?
  • a)
    50 cm
  • b)
    80 cm
  • c)
    40 cm
  • d)
    70 cm
Correct answer is option 'A'. Can you explain this answer?

Pranab Kapoor answered
Understanding the Metre Bridge Experiment
In a metre bridge experiment, the balance point (null point) indicates the equality of the ratios of resistances. Here’s a detailed explanation of how to determine the new null point when balancing different resistances.
Initial Setup
- A null point is found at 20 cm from one end of the wire when resistance X is balanced against resistance Y.
- This can be expressed as:
(X / Y) = (Length from one end / Length from the other end)
Thus, (X / Y) = (20 cm / 80 cm) = 1/4.
- Given that X < y,="" we="" can="" infer="" that="" x="">
New Resistance Balancing
- Now, we are tasked with balancing a resistance of 4X against Y.
- The new equation becomes (4X / Y) = (Length from one end / Length from the other end).
Calculating New Null Point
- From the previous relationship, we know that X = (1/4)Y, therefore:
4X = 4 * (1/4)Y = Y.
- This implies that when balancing 4X against Y, they are equal.
- Hence, the new null point would be halfway along the bridge length, which is at 50 cm from one end.
Conclusion
- Therefore, when balancing a resistance of 4X against Y, the new null point will be at 50 cm from the original end.
- Thus, the correct answer is option a) 50 cm.

A student measures the terminal potentialdifference (V) of a cell (of emf E and internalresistance r) as a function of the current (I)flowing through it. The slope and intercept, ofthe graph between V and I, then, respectively,equal: [2009]
  • a)
    – r and E
  • b)
    r and – E
  • c)
    – E and r
  • d)
    E and – r
Correct answer is option 'A'. Can you explain this answer?

Madhavan Patel answered
The slope of the graph between V and I represents the internal resistance of the cell, r. This is because according to Ohm's Law, V = E - Ir, where E is the emf of the cell. Rearranging this equation, we get Ir = E - V, which shows that the current I is directly proportional to the difference between the emf and the terminal potential difference. Therefore, the slope of the graph is equal to the internal resistance, r.

The intercept of the graph represents the emf of the cell, E. This is because when the current I is zero, the equation V = E - Ir simplifies to V = E, meaning that the terminal potential difference is equal to the emf. Therefore, the intercept of the graph is equal to the emf, E.

A cell can be balanced against 110 cm and 100cm of potentiometer wire, respectively with andwithout being short circuited through aresistance of 10Ω. Its internal resistance is [2008]
  • a)
    1.0 ohm
  • b)
    0.5 ohm
  • c)
    2.0 ohm
  • d)
    zero
Correct answer is option 'A'. Can you explain this answer?

Krish Khanna answered
Here   hence the lengths 110 cm
and 100 cm are interchanged.
Without being short-circuited through R,
only the battery E is balanced.
 
When R is connected across E, Ri, V/L, l2
Dividing (i) by (ii), we get
or, 100 R + 100 r = 110 R
or, 10 R = 100 r

The ratio of the resistance of conductor at temperature 15°C to its resistance at temperature 37.5°C is 4:5 . The temperature coefficient of resistance of the conductor is -
  • a)
     1/20 ºC-1
  • b)
     1/50 ºC-1
  • c)
    1/80 ºC-1
  • d)
     1/75 ºC-1
Correct answer is option 'D'. Can you explain this answer?

Rohan Pillai answered
Understanding Resistance and Temperature Relationship
The resistance of a conductor changes with temperature, which can be described by the formula:
R(T) = R0(1 + α(T - T0))
Where:
- R(T) = Resistance at temperature T
- R0 = Resistance at a reference temperature T0
- α = Temperature coefficient of resistance
- T = New temperature
- T0 = Reference temperature
Given Data
- Resistance ratio at 15°C and 37.5°C is 4:5.
- Let R1 = Resistance at 15°C and R2 = Resistance at 37.5°C.
From the ratio, we can express it as:
R1/R2 = 4/5
This leads to:
R2 = (5/4) R1
Applying the Resistance Formula
Using the resistance formula for both temperatures:
R1 = R0(1 + α(15 - T0))
R2 = R0(1 + α(37.5 - T0))
Substituting R2 into the resistance ratio:
R0(1 + α(15 - T0)) / R0(1 + α(37.5 - T0)) = 4/5
This simplifies to:
(1 + α(15 - T0)) / (1 + α(37.5 - T0)) = 4/5
Solving for Temperature Coefficient (α)
Cross-multiplying gives:
5(1 + α(15 - T0)) = 4(1 + α(37.5 - T0))
Expanding both sides:
5 + 5α(15 - T0) = 4 + 4α(37.5 - T0)
Rearranging leads to:
α(5(15 - T0) - 4(37.5 - T0)) = -1
Solving for α leads us to find:
α = -1 / (5(15 - T0) - 4(37.5 - T0))
When T0 is taken as 0°C, we find:
α = -1/75 °C^-1
Conclusion
The temperature coefficient of resistance of the conductor is indeed -1/75 °C^-1, confirming that option 'D' is the correct answer.

A galvanometer of resistance 50 Ω is connectedto battery of 3V along with a resistance of 2950 Ωin series. A full scale deflection of 30 divisions isobtained in the galvanometer. In order to reducethis deflection to 20 divisions, the resistance inseries should be [2008]
  • a)
    5050 Ω
  • b)
    5550 Ω
  • c)
    6050 Ω
  • d)
    4450 Ω
Correct answer is option 'D'. Can you explain this answer?

Nilanjan Chakraborty answered
Total internal resistance = (50+2950)Ω
= 3000 Ω
Emf of the cell, ε = 3V
Current for full scale deflection of 30
divisions is 1.0 mA.
∴ Current for a deflection of 20 divisions
Let the resistance be x Ω. Then
= 4500 Ω
But the resistance of the galvanometer
is 50Ω
∴ Resistance to be added
= (4500 –50) Ω= 4450 Ω
 

In producing chlorine by electrolysis 100 kWpower at 125 V is being consumed. How muchchlorine per minute is liberated? (E.C.E. ofchlorine is 0.367×10–6 kg / C) [2010]
  • a)
    1.76 × 10–3 kg
  • b)
    9.67 × 10. kg
  • c)
    17.61 × 10–3 kg
  • d)
    3.67 × 10–3 kg
Correct answer is option 'C'. Can you explain this answer?

Srishti Chavan answered
Calculation:
Given:
Power = 100 kW = 100,000 W
Voltage = 125 V
Charge of 1 mole of electrons (1 F) = 96500 C
E.C.E. of chlorine = 0.367 × 10⁻⁶ kg / C

Step 1: Calculate the charge consumed
Power (P) = Voltage (V) × Current (I)
Current (I) = P / V = 100,000 W / 125 V = 800 A

Step 2: Calculate the charge consumed per second
Charge (Q) = Current (I) × Time (t)
As power is given in kW and we need to find the amount per minute, convert time to seconds.
Time (t) = 1 minute = 60 seconds
Charge per second = 800 A × 60 s = 48,000 C

Step 3: Calculate the amount of chlorine liberated per second
Amount of chlorine liberated per second = E.C.E. of chlorine × Charge per second
= 0.367 × 10⁻⁶ kg / C × 48,000 C
= 17.616 × 10⁻³ kg

Step 4: Calculate the amount of chlorine liberated per minute
Amount of chlorine liberated per minute = Amount per second × 60
= 17.616 × 10⁻³ kg/s × 60 s
= 17.616 × 10⁻³ kg/min
= 17.61 × 10⁻³ kg/min
Therefore, the amount of chlorine liberated per minute is 17.61 × 10⁻³ kg, which is option C.

The sensitivity of the potentiometer can be increased by:
  • a)
    increasing the length of potentiometer wire.
  • b)
    increasing the e.m.f. of primary cell.
  • c)
    decreasing the length of potentiometer wire.
  • d)
    increasing the potential gradient.
Correct answer is option 'A'. Can you explain this answer?

Anjali Reddy answered
Explanation:A potentiometer is considered to be sensitive if the potential gradient dV/dl is low. Such a potentiometer can measure very small changes in potential difference. Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity. increasing the emf of the primary cell and by decreasing the length, potential gradient increases.

 If the length of the filament of a heater is reduced by 10% the power of the heater will
  • a)
    increase by about 11%
  • b)
    increase by about 19%
  • c)
    decrease by about 29%
  • d)
    Increase by about 9%
Correct answer is option 'A'. Can you explain this answer?

Top Rankers answered
Power P= V2/R​
If length is reduced by 10% then, new resistance of filament will be R′.
R′ = R−10% of R
R′ = 0.9R
Now new power of heater is P2​
P2​ = V2​​/R′ = V2/0.9R ​ ​= 1.1 P
% increase powe r = 11%

In the figure, voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are
  • a)
    125 V,3 A
  • b)
    100 V,4 A
  • c)
    120 V,4 A
  • d)
    120 V,3 A
Correct answer is option 'B'. Can you explain this answer?

Ambition Institute answered
Resistors 20Ω,100Ω and 25Ω will be in parallel. Their equivalent is 10Ω.


p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.

The internal resistance of a 2.1 V cell which givesa current of 0.2 A through a resistance of 10 Ω is[NEET 2013]
  • a)
    0.5 Ω
  • b)
    0.8 Ω
  • c)
    1.0 Ω
  • d)
    0.2 Ω
Correct answer is option 'A'. Can you explain this answer?

Abhiram Nair answered
Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1
⇒ 2.1 = 0.2r +2 ⇒ r = 1/2 = 0.5Ω

A wire of a certain material is stretched slowlyby ten per cent. Its new resistance and specificresistance become respectively: [2008]
  • a)
    1.2 times, 1.3 times
  • b)
    1.21 times, same
  • c)
    both remain the same
  • d)
    1.1 times, 1.1 times
Correct answer is option 'B'. Can you explain this answer?

Dipanjan Mehta answered
Resistance of a wire is given by R  l/a
If the length is increased by 10% then new
In that case, area of cross-section of wire
would decrease by 10%
∴ New area of cross-section
Thus the new resistance increases by 1.21
times. The specific resistance (resistivity)
remains unchanged as it depends on the
nature of the material of the wire.

If power dissipated in the 9-Ω resistor in the circuit shown is 36 watt, the potential difference across the 2-Ω resistor is [2011]
  • a)
    4 volt
  • b)
    8 volt
  • c)
    10 volt
  • d)
    2 volt
Correct answer is option 'C'. Can you explain this answer?

Anand Jain answered
We have, 
Current passing through the 9Ω resistor is
The 9Ω and 6Ω resistors are in parallel,
therefore
where i is the current delivered by the
battery.
Thus, potential difference across 2Ω
resistor is
V = iR
= 5 × 2
= 10V

A current of 3 amp flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5-Ω resistor is: [2008]
  • a)
    4 watt
  • b)
    2 watt
  • c)
    1 watt
  • d)
    5 watt
Correct answer is option 'D'. Can you explain this answer?

Rajeev Sharma answered
Clearly, 2Ω, 4Ω and ( 1 + 5) Ωresistors are
in parallel. Hence, potential difference is
same across each of them.
∴ I1 × 2 = I2 × 4 = I3 × 6
Given I1 = 3A ∴ I1 × 2 = I3 × 6
Given I1 = 3A.
∴ I1 × 2 = I3 × 6 provides
Now, the potential across the 5Ω resistor is
V = I3 × 5 = 1 × 5 = 5V.
∴ the power dissipated in the 5Ωresistor

In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will [2006]
  • a)
    flow in the direction which will be decidedby the value of V
  • b)
    be zero
  • c)
    flow from B to A
  • d)
    flow from A to B
Correct answer is option 'C'. Can you explain this answer?

Kajal Bose answered
Current will flow from B to A
Potential drop over the resistance CA will be
more due to higher value of resistance. So
potential at A will be less as compared with at
B. Hence, current will flow from B to A.

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths ℓ1 cm and ℓ2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to             [2010]
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Rohan Unni answered
(i) When key between the terminals 1
and 2 is plugged in,
P.D. across R = IR = k l1
⇒ R = k l1 as I = 1A
(ii) When key between terminals 1 and 3 is
plugged in,
P.D. across (X + R) = I(X + R) = k l2
⇒ X + R = k l2
∴ X = k (l2 – l1)
∴ R = kl1 and X = k (l2 – l1)

In the circuit shown, if the 10 Ω resistance is replaced by 20 Ω, then the amount of current drawn from the battery will be
  • a)
    10 A
  • b)
    4 A
  • c)
    8 A
  • d)
    2 A
Correct answer is option 'B'. Can you explain this answer?

Lead Academy answered
The network is a Wheatstone bridge. It’s balanced because
​,
so the potential at the top and bottom junctions is the same. Hence no current flows through the middle resistor, whatever its value (10 Ω or 20 Ω).
So the equivalent seen by the battery is just the two series paths in parallel:
With the original current 4 A, the battery voltage is
Replacing 10 Ω with 20 Ω doesn’t change Req​, so the current drawn remains

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