CBSE 2025, Set 3

Question

Calculate the value of the current passing through the battery in the given circuit diagram.

![Wheatstone Bridge](https://www.edvaya.com/images/cbse/physics/2025/set3/q17-wheatstone-bridge-resistor-network.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. Redrawing the circuit reveals it is a Wheatstone bridge.

The ratio of the arms is:
$$ \frac{10\Omega}{20\Omega} = \frac{20\Omega}{40\Omega} = \frac{1}{2} $$

Since the bridge is balanced, no current flows through the central $5\Omega$ resistor.

The equivalent resistance of the upper branch:
$$R_{upper} = 10\Omega + 20\Omega = 30\Omega$$

The equivalent resistance of the lower branch:
$$R_{lower} = 20\Omega + 40\Omega = 60\Omega$$

The total equivalent resistance $R_{eq}$ is the parallel combination of $30\Omega$ and $60\Omega$:
$$ R_{eq} = \frac{30 	imes 60}{30 + 60} $$
$$ R_{eq} = \frac{1800}{90} = 20\Omega $$

The total current passing through the battery is:
$$ I = \frac{V}{R_{eq}} $$
$$ I = \frac{6	ext{ V}}{20\Omega} = \frac{3}{10}	ext{ A} = \boxed{0.3	ext{ A}} $$

> Always check for a balanced Wheatstone bridge ($P/Q = R/S$) to simplify complex resistor networks.

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