A current of 0.1 ,A flows through a 25 ,Omega | vedclass

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(C) The circuit consists of a $25 \,\Omega$ resistor in series with a parallel combination of $20 \,\Omega$ and $60 \,\Omega$ resistors. First,calcula

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$0.3$

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(C) The circuit consists of a $25 \,\Omega$ resistor in series with a parallel combination of $20 \,\Omega$ and $60 \,\Omega$ resistors. First,calculate the equivalent resistance of the parallel part: $R_p = \frac{20 	imes 60}{20 + 60} = \frac{1200}{80} = 15 \,\Omega$. Now,the branch containing the $25 \,\Omega$ resistor has a total resistance of $R_{branch} = 25 \,\Omega + 15 \,\Omega = 40 \,\Omega$. The potential difference across this branch is $V_{AB} = I 	imes R = 0.1 \,A 	imes 40 \,\Omega = 4 \,V$. This same potential difference $V_{AB}$ is applied across the $20 \,\Omega$ resistor on the right. Thus,the current through the $20 \,\Omega$ resistor is $I_{20} = \frac{V_{AB}}{20 \,\Omega} = \frac{4 \,V}{20 \,\Omega} = 0.2 \,A$. According to Kirchhoff's junction rule at node $A$,the total current flowing through the $80 \,\Omega$ resistor is the sum of the currents in the two branches: $I_{total} = 0.1 \,A + 0.2 \,A = 0.3 \,A$.

$A$ current of $0.1 \,A$ flows through a $25 \,\Omega$ resistor as shown in the circuit diagram. The current in the $80 \,\Omega$ resistor is ........... $A$.

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