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 $R_p$ is given

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) First,simplify the circuit. The resistors $R_3 = 60\, \Omega$ and $R_4 = 30\, \Omega$ are in parallel. Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{60} + \frac{1}{30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}$,so $R_p = 20\, \Omega$.This $R_p$ is in series with $R_5 = 30\, \Omega$. The equivalent resistance of this branch is $R_{branch} = 20 + 30 = 50\, \Omega$.Now,this branch is in parallel with $R_2 = 50\, \Omega$. The equivalent resistance of this parallel combination is $R_{eq}' = \frac{50 	imes 50}{50 + 50} = 25\, \Omega$.Total resistance of the circuit is $R_{total} = R_1 + R_{eq}' = 50 + 25 = 75\, \Omega$.The total current from the battery is $I = \frac{V}{R_{total}} = \frac{3}{75} = \frac{1}{25}\, A = 0.04\, A$.This current $I$ flows through $R_1$ and then splits between $R_2$ and the branch containing $(R_3, R_4, R_5)$. Since $R_2 = 50\, \Omega$ and the branch resistance is $50\, \Omega$,the current splits equally: $I_2 = I_{branch} = \frac{I}{2} = \frac{1}{50}\, A = 0.02\, A$.The voltage across the parallel combination of $R_3$ and $R_4$ is $V_{p} = I_{branch} 	imes R_p = \frac{1}{50} 	imes 20 = \frac{2}{5}\, V = 0.4\, V$. 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 ohm and the battery is assumed ideal with $emf$ equal to $3\, V$. The voltage across the resistance $R_4$ is ................ $V$.

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