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

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(A) Let the potential at the node between $R_1$,$R_2$,and the $10\,\Omega$ resistor be $V$. The potential at the bottom wire is $0\,V$. The potential

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the potential at the node between $R_1$,$R_2$,and the $10\,\Omega$ resistor be $V$. The potential at the bottom wire is $0\,V$. The potential at the top wire after $R_1$ is $69\,V$. First,consider the rightmost branch with the $20\,\Omega$ resistor where a current of $1\,A$ flows. The voltage across this branch is $V = I 	imes R = 1\,A 	imes 20\,\Omega = 20\,V$. Thus,the potential at the node is $20\,V$. Now,for the branch with $R_2$,the current is $0.5\,A$ and the voltage across it is $20\,V$. Therefore,$R_2 = V / I = 20\,V / 0.5\,A = 40\,\Omega$. Next,consider the total current flowing from the battery. The current through $R_2$ is $0.5\,A$. The current through the $10\,\Omega$ resistor is $I_{10} = V / 10\,\Omega = 20\,V / 10\,\Omega = 2\,A$. The current through the $20\,\Omega$ resistor is $1\,A$. Total current $I_{total} = 0.5\,A + 2\,A + 1\,A = 3.5\,A$. This current flows through $R_1$. The voltage drop across $R_1$ is $69\,V - 20\,V = 49\,V$. So,$R_1 = 49\,V / 3.5\,A = 14\,\Omega$. Thus,$R_1 = 14\,\Omega$ and $R_2 = 40\,\Omega$.

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

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