In the figure shown below, a resistance of 150.4 \Omega is connected in series to an ammeter A of resistance 240 \Omega . A shunt resistance of 10 \Omega is connected in parallel with the ammeter. The reading of the ammeter is \underline\hspace2cm mA .

Current Electricity PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q71 with Solution

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Correct answer: 0.5

Interpret the circuit

An ammeter of resistance $240\,\Omega$ is shunted by a resistance of $10\,\Omega$ . This parallel combination is in series with a resistance of $150.4\,\Omega$ .

We need the current through the ammeter branch.

Find the equivalent resistance of the ammeter-shunt parallel combination

The ammeter resistance is $R_A = 240\,\Omega$

The shunt resistance is $R_s = 10\,\Omega$

Their parallel equivalent is

= \frac{240 \times 10}{240+10} = \frac{2400}{250} = 9.6\,\Omega$$ --- 3. **Find the total resistance of the circuit** This parallel combination is in series with $150.4\,\Omega$, so $$R_{\text{total}} = 150.4 + 9.6 = 160\,\Omega$$ --- 4. **Use the supply voltage from the figure** From the figure, the source voltage is $2\,\text{V}$. Hence total current in the series circuit is $$I = \frac{V}{R_{\text{total}}} = \frac{2}{160} = 0.0125\,\text{A} = 12.5\,\text{mA}$$ This $12.5\,\text{mA}$ enters the parallel combination. --- 5. **Find the current through the ammeter branch** Using current division, current through the ammeter is $$I_A = I\cdot \frac{R_s}{R_A + R_s}$$ $$I_A = 12.5\,\text{mA} \cdot \frac{10}{240+10} = 12.5\,\text{mA} \cdot \frac{10}{250} = 12.5\,\text{mA} \cdot 0.04 = 0.5\,\text{mA}$$ So the ammeter reading is $$\boxed{0.5\,\text{mA}}$$ --- 6. **Compare with stored correct answer** Stored correct answer: $5$ Our derived answer is $0.5$ mA, not $5$ mA. So the stored answer appears to be incorrect, unless the source voltage in the figure was actually $20\,\text{V}$ instead of $2\,\text{V}$. For a $20\,\text{V}$ source, the ammeter reading would indeed be $5$ mA.

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View written solutionFree

Correct answer: 0.5

Interpret the circuit

An ammeter of resistance $240\,\Omega$ is shunted by a resistance of $10\,\Omega$ . This parallel combination is in series with a resistance of $150.4\,\Omega$ .

We need the current through the ammeter branch.

Find the equivalent resistance of the ammeter-shunt parallel combination

The ammeter resistance is $R_A = 240\,\Omega$

The shunt resistance is $R_s = 10\,\Omega$

Their parallel equivalent is

= \frac{240 \times 10}{240+10} = \frac{2400}{250} = 9.6\,\Omega$$ --- 3. **Find the total resistance of the circuit** This parallel combination is in series with $150.4\,\Omega$, so $$R_{\text{total}} = 150.4 + 9.6 = 160\,\Omega$$ --- 4. **Use the supply voltage from the figure** From the figure, the source voltage is $2\,\text{V}$. Hence total current in the series circuit is $$I = \frac{V}{R_{\text{total}}} = \frac{2}{160} = 0.0125\,\text{A} = 12.5\,\text{mA}$$ This $12.5\,\text{mA}$ enters the parallel combination. --- 5. **Find the current through the ammeter branch** Using current division, current through the ammeter is $$I_A = I\cdot \frac{R_s}{R_A + R_s}$$ $$I_A = 12.5\,\text{mA} \cdot \frac{10}{240+10} = 12.5\,\text{mA} \cdot \frac{10}{250} = 12.5\,\text{mA} \cdot 0.04 = 0.5\,\text{mA}$$ So the ammeter reading is $$\boxed{0.5\,\text{mA}}$$ --- 6. **Compare with stored correct answer** Stored correct answer: $5$ Our derived answer is $0.5$ mA, not $5$ mA. So the stored answer appears to be incorrect, unless the source voltage in the figure was actually $20\,\text{V}$ instead of $2\,\text{V}$. For a $20\,\text{V}$ source, the ammeter reading would indeed be $5$ mA.

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