In the given figure,R_1=10 Omega, R_2=8 Omega | vedclass

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(D) From the circuit diagram,resistors $R_2, R_3,$ and $R_4$ are connected in parallel.The equivalent resistance $R_p$ of the parallel combination is

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$12 \Omega$ and $1 	ext{ A}$

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(D) From the circuit diagram,resistors $R_2, R_3,$ and $R_4$ are connected in parallel.The equivalent resistance $R_p$ of the parallel combination is given by:$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} = \frac{1+2+1}{8} = \frac{4}{8} = \frac{1}{2} \Omega^{-1}$.Thus,$R_p = 2 \Omega$.This parallel combination is in series with resistor $R_1$. Therefore,the total equivalent resistance $R_{eq}$ of the circuit is:$R_{eq} = R_1 + R_p = 10 \Omega + 2 \Omega = 12 \Omega$.The current $I$ supplied by the battery is given by Ohm's law:$I = \frac{V}{R_{eq}} = \frac{12 	ext{ V}}{12 \Omega} = 1 	ext{ A}$.

In the given figure,$R_1=10 \Omega, R_2=8 \Omega, R_3=4 \Omega$ and $R_4=8 \Omega$. The battery is ideal with an emf of $12 \text{ V}$. Find the equivalent resistance of the circuit and the current supplied by the battery.

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