In the figure shown,what is the current (in A | vedclass

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(C) First,observe the circuit. The resistors $R_3, R_4,$ and $R_5$ are in series. Their equivalent resistance is $R_s = R_3 + R_4 + R_5 = 20 + 5 + 25

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$9/32$

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(C) First,observe the circuit. The resistors $R_3, R_4,$ and $R_5$ are in series. Their equivalent resistance is $R_s = R_3 + R_4 + R_5 = 20 + 5 + 25 = 50\,\Omega$.This $R_s$ is in parallel with $R_2 = 10\,\Omega$. The equivalent resistance of this parallel combination is $R_p = \frac{R_2 	imes R_s}{R_2 + R_s} = \frac{10 	imes 50}{10 + 50} = \frac{500}{60} = \frac{25}{3}\,\Omega$.Now,the total equivalent resistance of the circuit is $R_{eq} = R_1 + R_p + R_6 = 15 + \frac{25}{3} + 30 = 45 + \frac{25}{3} = \frac{135 + 25}{3} = \frac{160}{3}\,\Omega$.The current drawn from the battery is $I = \frac{E}{R_{eq}} = \frac{15}{160/3} = \frac{15 	imes 3}{160} = \frac{45}{160} = \frac{9}{32}\,A$.

In the figure shown,what is the current (in Ampere) drawn from the battery? You are given $R_1 = 15\,\Omega$,$R_2 = 10\,\Omega$,$R_3 = 20\,\Omega$,$R_4 = 5\,\Omega$,$R_5 = 25\,\Omega$,$R_6 = 30\,\Omega$,and $E = 15\,V$.

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