Electric Current

61. Two long wires each carrying a d.c. current in the same direction are

Two long wires each carrying a d.c. current in the same direction are placed close to each other. Which one of the following statements is correct?

[amp_mcq option1=”The wires will attract each other” option2=”The wires will repel each other” option3=”There will be no force between the wires” option4=”There will be a force between the wires only at the moment when the current is switched ON or OFF” correct=”option1″]

This question was previously asked in
UPSC NDA-2 – 2015
Download PDFAttempt Online
When two parallel wires carry electric currents in the same direction, they produce magnetic fields that interact in such a way as to cause an attractive force between the wires.
This is a fundamental principle of electromagnetism: Parallel currents in the same direction attract, and parallel currents in opposite directions repel.
The force per unit length between two parallel wires carrying currents I₁ and I₂ separated by a distance r is given by F/L = (μ₀ * I₁ * I₂) / (2 * π * r), where μ₀ is the permeability of free space. The direction of the force depends on the relative direction of the currents.

62. Consider the following electric circuit : [Circuit diagram is present

Consider the following electric circuit :
[Circuit diagram is present in the image]
The current in the above electric circuit is :

[amp_mcq option1=”1 A” option2=”(10/15) A” option3=”2 A” option4=”1.5 A” correct=”option2″]

This question was previously asked in
UPSC NDA-1 – 2024
Download PDFAttempt Online
The current in the electric circuit is (10/15) A.
The circuit consists of a 10V battery connected to two resistors, 5 ohms and 10 ohms, in series. For resistors connected in series, the total resistance (R_total) is the sum of individual resistances. R_total = 5 Ω + 10 Ω = 15 Ω. According to Ohm’s Law, the current (I) flowing through the circuit is given by V = I * R_total, where V is the voltage. Therefore, I = V / R_total = 10 V / 15 Ω = (10/15) A.
The fraction (10/15) can be simplified to (2/3) A. The options provide the unsimplified fraction (10/15) A as option B. If the resistors were in parallel, the total resistance would be calculated differently (1/R_total = 1/R1 + 1/R2).

63. Which one of the following is the best shape of a solid metal rod to f

Which one of the following is the best shape of a solid metal rod to form the top end of a lightning conductor ?

[amp_mcq option1=”Figure (a)” option2=”Figure (b)” option3=”Figure (c)” option4=”Figure (d)” correct=”option1″]

This question was previously asked in
UPSC NDA-1 – 2024
Download PDFAttempt Online
The best shape for the top end of a lightning conductor is a sharp point (spike).
A sharp point concentrates the electric field lines from a charged cloud. This high electric field at the tip promotes corona discharge, which is a continuous leakage of charge from the conductor into the air. This discharge helps to neutralize the surrounding air and reduce the potential difference between the cloud and the ground, thereby reducing the likelihood of a direct lightning strike to the protected structure.
Blunt or rounded ends (like options b, c, and d) are less effective at concentrating the electric field and facilitating corona discharge compared to a sharp point. Modern lightning protection systems sometimes use other designs, but traditionally, the sharp point has been considered superior for this purpose.

64. If the current through an electrical machine running on direct current

If the current through an electrical machine running on direct current is 15 A and the machine runs for 10 minutes, the charge that passes through the machine during this time is :

[amp_mcq option1=”1.50 C” option2=”150 C” option3=”900 C” option4=”9000 C” correct=”option4″]

This question was previously asked in
UPSC NDA-1 – 2024
Download PDFAttempt Online
The charge (Q) that passes through a machine is given by the product of the current (I) and the time (t) for which the current flows: Q = I * t.
The current is given in Amperes (A), and the time must be in seconds (s) to get the charge in Coulombs (C). The given time is 10 minutes, which is 10 * 60 = 600 seconds. The current is 15 A.
Calculating the charge: Q = 15 A * 600 s = 9000 Coulombs. This represents the total amount of electric charge that flows through the electrical machine over the 10-minute period.

65. An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω

An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω.
The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to :
where value of V₂ = 5 V. The internal resistances of both the batteries are 0.1 Ω. The current through the resistance R is about :

[amp_mcq option1=”1.0 mA” option2=”1.2 mA” option3=”3.0 mA” option4=”5.0 mA” correct=”option3″]

This question was previously asked in
UPSC NDA-1 – 2024
Download PDFAttempt Online
The first part of the question describes a circuit with a battery V1=1V, resistor R=1000Ω, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V/R = 1V/1000Ω = 1mA, consistent with the description.
The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1Ω. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω) are connected in parallel across A and B, and the resistor R (1000Ω) is also connected between A and B.
In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A.
Current through R: I_R = V_A / R
Current from V1 source: I_1 = (V1 – V_A) / r1 = (1 – V_A) / 0.1
Current from V2 source: I_2 = (V2 – V_A) / r2 = (5 – V_A) / 0.1
Applying KCL at node A (sum of currents leaving A is zero):
I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let’s assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B.
Currents leaving A: I_R = V_A / R, I_1 = V_A – V1 / r1? No. This implies V_A is higher than the source terminal.
Let’s redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B).
Using superposition or combining parallel sources:
Equivalent voltage source V_eq = (V1/r1 + V2/r2) / (1/r1 + 1/r2)
Equivalent internal resistance r_eq = 1 / (1/r1 + 1/r2)
V_eq = (1V/0.1Ω + 5V/0.1Ω) / (1/0.1Ω + 1/0.1Ω) = (10 A + 50 A) / (10 S + 10 S) = 60 A / 20 S = 3 V.
r_eq = 1 / (10 S + 10 S) = 1 / 20 S = 0.05 Ω.
Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05Ω connected across R=1000Ω.
The voltage across R is V_AB = V_eq * R / (R + r_eq) = 3V * 1000Ω / (1000Ω + 0.05Ω) = 3V * 1000 / 1000.05.
V_AB ≈ 3V.
The current through R is I_R = V_AB / R = (3V * 1000 / 1000.05) / 1000Ω = 3V / 1000.05Ω.
I_R ≈ 3 / 1000 = 0.003 A = 3 mA.
Calculating precisely: 3 / 1000.05 ≈ 0.00299985 A ≈ 2.99985 mA.
This is very close to 3.0 mA.

– The first part of the question establishes the context: R=1000Ω, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source.
– The second part adds V2=5V and internal resistances 0.1Ω for *both* batteries.
– Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence.
– When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq.
– Formula for parallel sources: V_eq = (Σ Vi/ri) / (Σ 1/ri) and r_eq = 1 / (Σ 1/ri).
– Calculate V_eq and r_eq for V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω).
– Calculate the current through R (1000Ω) when connected across the equivalent source.
– Calculation gives current ≈ 3 mA.
If the batteries were in series, the total voltage would be 6V (aiding) or 4V (opposing). Total internal resistance would be 0.2Ω. Current through 1000Ω would be approximately 6mA or 4mA. Neither of these is a very close match to the options (3.0 mA or 5.0 mA), whereas 3mA is a precise match for the parallel configuration calculation result of ~2.99985 mA. This strongly supports the parallel connection interpretation.

66. An infinite combination of resistors, each having resistance R = 4 Ω,

An infinite combination of resistors, each having resistance R = 4 Ω, is given below. What is the net resistance between the points A and B ? (Each resistance is of equal value, R = 4)

[amp_mcq option1=”0 Ω” option2=”2 + 2√5 Ω” option3=”2 + √5 Ω” option4=”∞ Ω” correct=”option2″]

This question was previously asked in
UPSC NDA-1 – 2024
Download PDFAttempt Online
This is an infinite ladder network of resistors. Let the equivalent resistance between points A and B be R_eq. Since the network is infinite and uniform, if we remove the first segment of the ladder (consisting of two horizontal resistors R and one vertical resistor R), the remaining infinite network to the right will have the same equivalent resistance R_eq. The circuit between A and B can be viewed as the first horizontal resistor R (on the top rail) in series with the parallel combination of the vertical resistor R and the equivalent resistance of the rest of the ladder (R_eq), and this combination is then in series with the first horizontal resistor R (on the bottom rail).
Wait, re-reading the visual interpretation and the standard ladder problem formulation: The equivalent resistance R_eq of the infinite ladder from point A to point B (across the first vertical rung) can be determined by considering that adding one more segment at the beginning does not change the total resistance. The segment added at the beginning consists of a horizontal R, a vertical R, and another horizontal R.
Let R_eq be the resistance of the infinite ladder starting from one junction pair (like A and B) to infinity. The structure is: A — R — J1 — …inf… ; B — R — K1 — …inf… ; J1 — R — K1.
The resistance between J1 and K1 onwards is also R_eq.
So, the resistance between A and B is R (A to J1) in series with the parallel combination of R (J1 to K1) and R_eq (J1 onwards), and this whole combination in series with R (K1 to B). This assumes A and B are the start points on the two rails *before* the first vertical resistor. This interpretation doesn’t match the diagram precisely, where A and B are connected to the points *before* the first horizontal resistors on the top and bottom rails respectively.

Let’s use the standard ladder network definition where A and B are the input points. The network structure repeats. Let R_eq be the equivalent resistance of the infinite network from points A and B.
Adding the first segment (R horizontal top, R vertical, R horizontal bottom) such that the rest of the infinite ladder connected after the first vertical resistor has resistance R_eq.
The total resistance R_eq is the resistance of the first horizontal top R, plus the parallel combination of the first vertical R and the resistance of the rest of the ladder. The rest of the ladder, starting from the points where the first vertical R connects to the rails, itself looks like the infinite ladder.
Let’s assume A and B are the input terminals, and the network extends to infinity. The resistance seen from A and B is R_AB. The first segment is horizontal R on top, horizontal R on bottom, and vertical R between them. This interpretation still doesn’t match the usual ladder structure formula derived above.

Let’s re-examine the formula derivation that matches option B. The formula R_eq = R * (1 + sqrt(5)) / 2 corresponds to an infinite ladder where *each segment* consists of a series resistance R on one rail, a series resistance R on the other rail, and a shunt resistance R between the rails. The resistance is calculated between corresponding points on the two rails at the beginning.
A — R — R — …
| |
R R
| |
B — R — R — …
Let R_eq be the resistance between A and B. If we add one more segment (R top, R bottom, R vertical) at the beginning, the resistance between the new input points should still be R_eq.
This structure is slightly different from the image. The image shows horizontal R, vertical R, horizontal R, vertical R…
A — R — J1
|
R
|
K1 — R — J2
|
R
|
K2 — …
B is connected to K1. This is not A and B being the input points of the ladder. A and B are connected as shown in the diagram.
Let the resistance of the infinite network starting from J1 and K1 onwards be R_inf.
The resistance between J1 and B is the series combination of R (J1 to K1) and R (K1 to B). So R_J1B = R + R = 2R.
The resistance between A and J1 is R.
The resistance from A to B is R (A to J1) in series with the parallel combination of R (vertical from J1) and the resistance of the rest of the network starting from K1. This is still confusing based on the diagram.

Let’s assume the question *implies* the standard infinite ladder network structure where A and B are the input terminals across the first vertical resistor, and the formula R_eq = R * (1 + sqrt(5)) / 2 applies to this structure.
In that case, with R = 4 Ω:
R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.
This matches option B. Given this is a standard physics/engineering problem, it’s highly likely the question intends this standard structure and formula, despite the potentially ambiguous labeling of A and B in the diagram relative to the text description (“between the points A and B”). Let’s proceed with this interpretation as it’s the only one yielding a matching option.

– The network is an infinite ladder structure made of identical resistors R.
– The key to solving infinite networks is to recognize that adding or removing a finite number of repeating units does not change the overall equivalent resistance of the infinite part.
– Let the equivalent resistance of the infinite ladder from a typical junction pair onwards be R_eq.
– By considering the first segment and the remaining infinite part, a relationship can be set up to solve for R_eq.
– For a standard ladder with series resistors R on rails and shunt resistors R between rails, the equivalent resistance R_eq satisfies R_eq = R + (R * R_eq) / (R + R_eq). This leads to R_eq^2 – R * R_eq – R^2 = 0, with the positive solution R_eq = R * (1 + sqrt(5)) / 2.
– Applying R = 4 Ω, R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.
The term (1 + sqrt(5)) / 2 is the golden ratio, often denoted by φ (phi). So, the equivalent resistance of such a ladder network is R * φ. Infinite networks are often solved using recurrence relations or by assuming the equivalent resistance of the infinite part is the same as the whole.

67. An electric bulb is rated as 220 V and 80 W. When it is operated on 11

An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be :

[amp_mcq option1=”80 W” option2=”60 W” option3=”40 W” option4=”20 W” correct=”option4″]

This question was previously asked in
UPSC NDA-1 – 2023
Download PDFAttempt Online
The power rating of an electric bulb is given by the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. The resistance of the bulb remains constant (assuming its temperature change is negligible for this calculation). First, calculate the resistance using the rated values: R = V_rated^2 / P_rated = (220 V)^2 / 80 W = 48400 / 80 = 605 Ω. When operated on 110 V, the power consumed is P_new = V_new^2 / R = (110 V)^2 / 605 Ω = 12100 / 605 = 20 W.
– The resistance of the bulb is a fixed property.
– Power is proportional to the square of the voltage when resistance is constant (P ∝ V^2).
– Since the voltage is halved (110V is half of 220V), the power will be reduced by a factor of (1/2)^2 = 1/4.
– New Power = Rated Power * (New Voltage / Rated Voltage)^2 = 80 W * (110 V / 220 V)^2 = 80 W * (1/2)^2 = 80 W * (1/4) = 20 W.
This calculation assumes the resistance R is constant, which is a reasonable approximation for this type of problem, although in reality, the resistance of a filament bulb changes with temperature.

68. What is the total resistance in the following circuit element ? [Image

What is the total resistance in the following circuit element ?
[Image of a circuit diagram with three resistors labelled R. Two are in parallel, and this combination is in series with the third resistor.]

[amp_mcq option1=”R/2″ option2=”3R” option3=”3R/2″ option4=”2R/3″ correct=”option3″]

This question was previously asked in
UPSC NDA-1 – 2023
Download PDFAttempt Online
The circuit element consists of three resistors, each with resistance R. Two of the resistors are connected in parallel, and this parallel combination is connected in series with the third resistor. To find the total resistance, we first calculate the equivalent resistance of the parallel part and then add it to the resistance of the series part.
– For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + …$
– For resistors in series, the equivalent resistance is the sum of individual resistances: $R_s = R_1 + R_2 + …$
Let the resistance of each resistor be R.
The two resistors in parallel have equivalent resistance $R_p$.
$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$.
$R_p = \frac{R}{2}$.
This parallel combination ($R_p$) is in series with the third resistor (R).
The total resistance $R_{total}$ is the sum of the series components:
$R_{total} = R_p + R = \frac{R}{2} + R = \frac{R + 2R}{2} = \frac{3R}{2}$.

69. In an electric circuit, a wire of resistance 10 Ω is used. If this wir

In an electric circuit, a wire of resistance 10 Ω is used. If this wire is stretched to a length double of its original value, the current in the circuit would become :

[amp_mcq option1=”half of its original value.” option2=”double of its original value.” option3=”one-fourth of its original value.” option4=”four times of its original value.” correct=”option3″]

This question was previously asked in
UPSC NDA-1 – 2023
Download PDFAttempt Online
The resistance of a wire depends on its material (resistivity), length, and cross-sectional area. When a wire is stretched, its length increases, and its cross-sectional area decreases, while its volume remains constant. This change in dimensions affects the resistance. The current in the circuit is inversely proportional to the resistance (assuming constant voltage).
– Resistance R = $\rho \frac{L}{A}$, where $\rho$ is resistivity, L is length, and A is cross-sectional area.
– Volume of the wire V = A × L. When stretched, the volume remains constant: $A’L’ = AL$.
– Ohm’s Law: Current I = Voltage V / Resistance R.
Let the original length be L and the original cross-sectional area be A. The original resistance is $R = \rho \frac{L}{A}$.
The wire is stretched to a new length $L’ = 2L$.
Since the volume remains constant, $A’L’ = AL$.
$A'(2L) = AL \Rightarrow A’ = \frac{A}{2}$.
The new resistance is $R’ = \rho \frac{L’}{A’} = \rho \frac{2L}{A/2} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A}$.
So, $R’ = 4R$. The resistance becomes four times the original resistance.
Assuming the voltage V across the circuit element (or the voltage source) is constant:
Original current $I = \frac{V}{R}$.
New current $I’ = \frac{V}{R’} = \frac{V}{4R} = \frac{1}{4} \left(\frac{V}{R}\right) = \frac{1}{4}I$.
The current in the circuit would become one-fourth of its original value.

70. A positive charge +q is placed at the centre of a hollow metallic sphe

A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. The electric field at a distance r from the centre is denoted by E. In this regard, which one of the following statements is correct ?

[amp_mcq option1=”E = 0 for a < r < b" option2="E = 0 for r < a" option3="E = q/4πε₀r for a < r < b" option4="E = q/4πε₀a for r < a" correct="option1"]

This question was previously asked in
UPSC NDA-1 – 2017
Download PDFAttempt Online
A positive charge +q is placed at the center of a hollow metallic sphere.
– For r < a (inside the cavity but outside the point charge): The electric field is due to the point charge +q. Using Gauss's Law with a spherical surface of radius r < a centered at +q, the enclosed charge is +q. So, E * 4πr² = q/ε₀, which gives E = q/(4πε₀r²). Thus, E is not zero for r < a. - For a < r < b (inside the metallic material): In electrostatic equilibrium, the electric field inside a conductor is always zero. The presence of the charge +q at the center induces a charge of -q on the inner surface of the metallic sphere (at r=a) and a charge of +q on the outer surface (at r=b). The induced charges arrange themselves such that the field inside the conductor material (a < r < b) is zero. - For r > b (outside the sphere): The total enclosed charge within a spherical surface of radius r > b centered at +q is +q (the original charge) + (-q on inner surface) + (+q on outer surface) = +q. Using Gauss’s Law, E * 4πr² = q/ε₀, so E = q/(4πε₀r²).
Based on this analysis, the statement E = 0 for a < r < b is correct.
This question tests fundamental concepts of electrostatics, specifically the electric field inside and outside conductors and due to point charges, and the effect of induced charges in conductors.
The electric field inside a conductor in electrostatic equilibrium is zero because free charges within the conductor redistribute themselves to cancel any external or internal electric field. This phenomenon is known as electrostatic shielding.
Page 1Page 2Page 3Page 4Page 5