CBSE 2025, Compartment, Set 1

Question

The figure shows a network of resistors connected across a battery. Find the current supplied by the battery of emf $5	ext{ V}$ and internal resistance $\frac{2}{3}  \Omega$.

![Wheatstone bridge circuit](https://www.edvaya.com/images/cbse/physics/2025/wheatstone-bridge-circuit.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** The given network matches the topology of a Wheatstone Bridge circuit. Let's check the ratio of resistances in the parallel arms. $$\frac{R_{AB}}{R_{AD}} = \frac{8 \Omega}{4 \Omega} = 2$$ $$\frac{R_{BC}}{R_{DC}} = \frac{2 \Omega}{1 \Omega} = 2$$ Since the bridge ratios are equal, it represents a perfectly **balanced Wheatstone bridge**. No current will flow through the central $3 \Omega$ resistor, so it can be safely removed from calculations. **Equivalent Resistance ($R$):** The upper branch series resistance $= 8 + 2 = 10 \Omega$ The lower branch series resistance $= 4 + 1 = 5 \Omega$ These two branches are in parallel. $$\frac{1}{R} = \frac{1}{10} + \frac{1}{5}$$ $$\frac{1}{R} = \frac{1+2}{10} = \frac{3}{10}$$ $$R = \frac{10}{3} \Omega$$ **Total Current ($I$):** Using Ohm's law, $I = \frac{E}{R + r}$ where internal resistance $r = \frac{2}{3} \Omega$. $$I = \frac{5}{\frac{10}{3} + \frac{2}{3}}$$ $$I = \frac{5}{\frac{12}{3}} = \frac{5}{4}$$ $$I = 	extbf{1.25 A}$$

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