Physics Q49 – JEE-MAIN 2025

Home Physics Current ElectricityQuestion #55

In the figure shown below, a resistance of

150.4 Ω

is connected in series to an ammeter A of resistance

240 Ω

. A shunt resistance of

10 Ω

is connected in parallel with the ammeter. The reading of the ammeter is _______ mA.

Enter Calculated Numerical Value:

Clue 1 of 3: Circuit Analysis

Identify the series and parallel combinations of resistors in the circuit to determine the overall circuit structure.

🧠 Full Solution Path

Step 1: Calculate the equivalent resistance of the parallel combination.✦ Active

The ammeter (resistance $R_{A} = 240 Ω$ ) and the shunt resistance ( $R_{s h} = 10 Ω$ ) are connected in parallel. Their equivalent resistance $R_{p}$ is calculated as:

R_{p} = \frac{R_{A} \times R_{s h}}{R_{A} + R_{s h}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 Ω

Step 2: Calculate the total current in the circuit.○ Expand

The main resistance ( $R_{1} = 150.4 Ω$ ) is in series with the parallel combination $R_{p}$ . The total equivalent resistance of the circuit is $R_{e q} = R_{1} + R_{p}$ . The total current $I_{t o t a l}$ drawn from the $20 V$ source is then found using Ohm's Law:

R_{e q} = 150.4 Ω + 9.6 Ω = 160 Ω

\Rightarrow I_{t o t a l} = \frac{V}{R_{e q}} = \frac{20 V}{160 Ω} = 0.125 A

Step 3: Determine the current through the ammeter.○ Expand

The total current $I_{t o t a l}$ splits between the ammeter and the shunt. The current through the ammeter $I_{A}$ can be found using the current divider rule. Finally, convert the current from Amperes to milliamperes.

I_{A} = I_{t o t a l} \times \frac{R_{s h}}{R_{A} + R_{s h}} = 0.125 A \times \frac{10 Ω}{240 Ω + 10 Ω} = 0.125 A \times \frac{10}{250} = 0.125 A \times \frac{1}{25} = 0.005 A

\Rightarrow I_{A} = 0.005 A \times 1000 \frac{m A}{A} = 5 m A

💡 Teacher's Secret Hint

Ensure you apply the current divider rule correctly, using the resistance of the *other* branch in the numerator.

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Physics Question 49 – JEE-MAIN 2025

Clue 1 of 3: Circuit Analysis

🧠 Full Solution Path