Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Two resistors of 200 $\Omega$ and 400 $\Omega$ are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 $\Omega$ resistance. The potential drop across the bulb is ________ V.
(1) 25
(2)
(3)
(4)
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Answer: Option (50)
Key idea: Series-Parallel Circuits and Voltage Division
A voltmeter with internal resistance of $x \Omega$ can be used to measure upto 20 V. In order to increase its measuring range to 30 V, the required modification is to______.
(1) connect resistor of $\frac{x}{2} \Omega$, in series with voltmeter.
(2) connect resistor of $\frac{x}{2} \Omega$, in parallel to voltmeter.
(3) connect a resistor of $x \Omega$ in series with voltmeter.
(4) connect resistor of $2x \Omega$ in parallel to voltmeter.
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Answer: Option (1)
Key idea: Voltmeter Range Extension using Series Multiplier Resistance
The voltmeter with resistance $x\Omega$ reads maximum 20 V, so maximum current through it is $I = \frac{20}{x}$ A
To measure 30 V, the voltmeter still reads 20 V (its max), and extra 10 V must drop across a series resistor
For range extension, connect a multiplier resistance $R_s$ in **series** with the voltmeter
Using Ohm's law: $V_s = I \times R_s$ gives $10 = \frac{20}{x} \times R_s$
The voltage and the current between $A$ and $B$ points shown in the circuit are ______.
[Figure: Circuit diagram with three parallel branches between points A and B. Top branch: 27 V source and 3 $\Omega$ resistor. Middle branch: 27 V source and 3 $\Omega$ resistor. Bottom branch: 14 V and 13 V sources in series with 3 $\Omega$ resistor leading to point B, then two 3 $\Omega$ resistors in series, followed by 27 V source and 3 $\Omega$ resistor back to point A]
(1) 24 V, 12 A
(2) 24 V, 4 A
(3) 18 V, 12 A
(4) 27 V, 4 A
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Refer to the figure given below, current between terminals $A$ and $B$ is ______. [Figure: Circuit diagram showing a network of resistors and voltage sources arranged in rows between terminals A and B. The circuit has three rows with 5V sources and 3Ω resistors in series in each branch, and additional 3Ω resistors at the bottom connecting A to B]
(1) 12.5
(2) 1.25
(3) 7.5
(4) 5
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Answer: Option (2)
Key idea: Parallel combination of cells with internal resistance, Kirchhoff's laws, equivalent EMF and resistance
Identify the circuit: Three parallel branches between A and B, each containing a 5V source and 3Ω resistor in series, plus an additional 3Ω resistor
For parallel branches with identical EMF and resistance: $E_{eq} = 5V$ (same for all branches)
Under steady state condition the potential difference across the capacitor in the circuit is _______ V.
[Figure: Circuit diagram showing a 2V battery connected to a network with a 2Ω resistor in series with two parallel branches - one containing a 2μF capacitor in series with a 4Ω resistor, and another containing a 6Ω resistor]
(1) 0.5
(2) 1.5
(3) 0
(4) 2
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Answer: Option (2)
Key idea: Capacitor in steady state acts as open circuit; no current flows through capacitor branch, so entire branch voltage appears across capacitor
In steady state, capacitor is fully charged and acts as **open circuit** (no current through that branch)
Current flows only through: $2\Omega$ resistor → $6\Omega$ resistor path
Total resistance: $R_{total} = 2\Omega + 6\Omega = 8\Omega$
When an external resistance of 5 $\Omega$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 $\Omega$ resistor is replaced by a 2 $\Omega$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is __________ $\Omega$.
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Answer: 1
Key idea: EMF and internal resistance of a cell using Ohm's law
For a cell with internal resistance $r$ and EMF $E$: $E = I(R + r)$
**Case 1:** With $R_1 = 5\Omega$ and $I_1 = 0.25$ A: $E = 0.25(5 + r) = 1.25 + 0.25r$
**Case 2:** With $R_2 = 2\Omega$ and $I_2 = 0.5$ A: $E = 0.5(2 + r) = 1 + 0.5r$
Equate both expressions: $1.25 + 0.25r = 1 + 0.5r$
Solve: $0.25 = 0.25r$, therefore $r = 1\Omega$
Verify: $E = 1.5$ V gives correct currents in both cases
The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance $R_S$ is 0.5 mA. The minimum value of the resistance of $R_S$, to ensure that the LED is not damaged is ______ k$\Omega$. [Figure: Circuit diagram with 5V battery, resistor $R_S$ in series, followed by parallel combination of resistor $R = 1 k\Omega$ and 2 mW LED]
(1) 6
(2) 2
(3) 4
(4) 5
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Refer to the circuit diagram given below. The heat generated across the $6 \Omega$ resistance in 100 second is $\frac{a}{100}$ J. The value of $a$ is _______. (Nearest integer)
[Figure: A circuit diagram showing a 2V battery connected to a 3Ω resistor in series, which then splits into two parallel branches - one with a 6Ω resistor and 3V battery in series, and another with a 4Ω resistor]
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Answer: 17305
Key idea: Kirchhoff's Laws, Heat dissipation in resistors, Complex DC circuits with multiple batteries
Apply Kirchhoff's Current Law: $I = I_1 + I_2$ where $I_1$ is current through $6\Omega$ and $I_2$ through $4\Omega$
Two cells of emfs 1 V and 2 V and internal resistance $2\,\Omega$ and $1\,\Omega$, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\frac{\alpha}{5}$ A. The value of $\alpha$ is __________.
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Answer: 3
Key idea: Cells in parallel with same and opposite polarities, equivalent EMF and internal resistance
For parallel cells with same polarity: $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{1/2 + 2/1}{1/2 + 1} = \frac{5/3}{3/2} = \frac{5}{3}$ V
The stored charge in the capacitor in steady state of the following circuit is __________ $\mu$C.
[Figure: Circuit diagram showing a 12V battery connected to a network of resistors (5$\Omega$, 4$\Omega$, 10$\Omega$, 12$\Omega$, 10$\Omega$, 4$\Omega$, 2$\Omega$, 2$\Omega$) and a 100$\mu$F capacitor arranged in a combination of series and parallel connections]
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Answer: 600
Key idea: Capacitor in DC circuit at steady state, Kirchhoff's laws, Voltage divider rule
In steady state, the capacitor acts as an open circuit (no current flows through it)
Identify the two nodes across which the capacitor is connected in the circuit
Use Kirchhoff's voltage law and voltage divider rule to find potential at each node
Calculate the potential difference: $V_C = |V_1 - V_2|$ across the capacitor
Use the formula $Q = CV$ where $C = 100\mu F$ to find the stored charge
