In the circuit shown in the given figure,the | vedclass

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(A) First,analyze the rightmost branch where a current of $1\,A$ flows through a $20\,\Omega$ resistor. The voltage drop across this resistor is $V =

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$14\,\Omega$ and $40\,\Omega$

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(A) First,analyze the rightmost branch where a current of $1\,A$ flows through a $20\,\Omega$ resistor. The voltage drop across this resistor is $V = I 	imes R = 1\,A 	imes 20\,\Omega = 20\,V$. This implies the node above the $20\,\Omega$ resistor is at $20\,V$ and the bottom node is at $0\,V$.Next,consider the $10\,\Omega$ resistor. The current through it is $I = V/R = 20\,V / 10\,\Omega = 2\,A$. Applying Kirchhoff's Current Law $(KCL)$ at the top node,the total current entering from the left branch must equal the sum of currents leaving through the $R_2$ branch,the $10\,\Omega$ branch,and the $20\,\Omega$ branch. Total current $I_{total} = 0.5\,A + 2\,A + 1\,A = 3.5\,A$.For resistor $R_2$,the voltage across it is $20\,V - 0\,V = 20\,V$. Given the current through it is $0.5\,A$,we have $R_2 = V/I = 20\,V / 0.5\,A = 40\,\Omega$.For resistor $R_1$,the voltage drop across it is $69\,V - 20\,V = 49\,V$. The current flowing through $R_1$ is $3.5\,A$. Thus,$R_1 = V/I = 49\,V / 3.5\,A = 14\,\Omega$.Therefore,$R_1 = 14\,\Omega$ and $R_2 = 40\,\Omega$.

In the circuit shown in the given figure,the resistances $R_1$ and $R_2$ are respectively:

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