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

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(A) Let the nodes be labeled as shown in the figure. The potential difference across the $20\,\Omega$ resistor is $V_{CD} = I 	imes R = 1\,A 	imes 2

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$14\,\Omega 	ext{ and } 40\,\Omega$

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(A) Let the nodes be labeled as shown in the figure. The potential difference across the $20\,\Omega$ resistor is $V_{CD} = I 	imes R = 1\,A 	imes 20\,\Omega = 20\,V$.Since the $20\,\Omega$ resistor,$10\,\Omega$ resistor,and $R_2$ are in parallel,the potential difference across each is the same.Thus,$V_{BE} = V_{CD} = 20\,V$.For resistor $R_2$,$V_{BE} = I_{R2} 	imes R_2 \Rightarrow 20\,V = 0.5\,A 	imes R_2 \Rightarrow R_2 = 40\,\Omega$.For the $10\,\Omega$ resistor,the current $i = V_{BE} / 10\,\Omega = 20\,V / 10\,\Omega = 2\,A$.The total current $i_0$ flowing through $R_1$ is the sum of currents in the parallel branches: $i_0 = 0.5\,A + 2\,A + 1\,A = 3.5\,A$.Applying Kirchhoff's Voltage Law in the outer loop: $69\,V - i_0 R_1 - V_{BE} = 0$.$69\,V - 3.5\,A 	imes R_1 - 20\,V = 0$.$3.5\,A 	imes R_1 = 49\,V$.$R_1 = 49 / 3.5 = 14\,\Omega$.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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