Question 13

In the given figure $$R_1 = 10\Omega, R_2 = 8\Omega, R_3 = 4\Omega$$ and $$R_4 = 8\Omega$$. Battery is ideal with emf $$12 \text{ V}$$. Equivalent resistance of the circuit and current supplied by battery are respectively :

$$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_4} + \frac{1}{R_3}$$

$$\implies \frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} + \frac{1}{4} = \frac{1}{2} \implies R_p = 2\ \Omega$$

$$R_{\text{eq}} = R_1 + R_p = 10 + 2 = 12\ \Omega$$

$$I = \frac{V}{R_{\text{eq}}} = \frac{12}{12} = 1\text{ A}$$

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