In the circuit shown below,the resistances ar | vedclass

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(A) First,simplify the circuit. The resistors $R_3 = 60\,\Omega$ and $R_4 = 30\,\Omega$ are in parallel. Their equivalent resistance $R_{34} = \frac{6

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$0.4$

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(A) First,simplify the circuit. The resistors $R_3 = 60\,\Omega$ and $R_4 = 30\,\Omega$ are in parallel. Their equivalent resistance $R_{34} = \frac{60 	imes 30}{60 + 30} = \frac{1800}{90} = 20\,\Omega$.This combination is in series with $R_5 = 30\,\Omega$,so the branch resistance is $R_{345} = 20 + 30 = 50\,\Omega$.This branch is in parallel with $R_2 = 50\,\Omega$,so the equivalent resistance of this part is $R_{2345} = \frac{50 	imes 50}{50 + 50} = 25\,\Omega$.Finally,$R_1 = 50\,\Omega$ is in series with this,so the total equivalent resistance is $R_{eq} = 50 + 25 = 75\,\Omega$.The total current from the battery is $I = \frac{V}{R_{eq}} = \frac{3}{75} = 0.04\,A$.The voltage across the parallel combination of $R_2$ and the branch $(R_3, R_4, R_5)$ is $V_p = I 	imes R_{2345} = 0.04 	imes 25 = 1\,V$.The current through the branch containing $R_3$ and $R_4$ is $I_{branch} = \frac{V_p}{R_{345}} = \frac{1}{50} = 0.02\,A$.The voltage across the parallel combination of $R_3$ and $R_4$ is $V_{34} = I_{branch} 	imes R_{34} = 0.02 	imes 20 = 0.4\,V$.Since $R_3$ and $R_4$ are in parallel,the voltage across $R_4$ is $0.4\,V$.

In the circuit shown below,the resistances are given in $\Omega$ and the battery is assumed ideal with an emf equal to $3\,V$. The voltage across the resistance $R_4$ is ................. $V$.

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