The current from the cell for the given circu | vedclass

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(A) The circuit consists of a bridge-like structure. Let the nodes be $A$ and $B$ across the $20 \, \Omega$ resistor. The upper branch has $10 \, \Ome

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The circuit consists of a bridge-like structure. Let the nodes be $A$ and $B$ across the $20 \, \Omega$ resistor. The upper branch has $10 \, \Omega$ and $6 \, \Omega$ in series,and the lower branch has $5 \, \Omega$ and $3 \, \Omega$ in series.This is a balanced Wheatstone bridge because $\frac{10}{5} = \frac{6}{3} = 2$. Thus,no current flows through the $20 \, \Omega$ resistor.The equivalent resistance of the upper branch is $10 + 6 = 16 \, \Omega$.The equivalent resistance of the lower branch is $5 + 3 = 8 \, \Omega$.These two branches are in parallel with the $16 \, \Omega$ resistor.Let $R_p$ be the equivalent resistance of the parallel combination of the $16 \, \Omega$ (upper branch),$8 \, \Omega$ (lower branch),and $16 \, \Omega$ (middle resistor).$\frac{1}{R_p} = \frac{1}{16} + \frac{1}{8} + \frac{1}{16} = \frac{1+2+1}{16} = \frac{4}{16} = \frac{1}{4} \, \Omega^{-1}$.So,$R_p = 4 \, \Omega$.The total resistance of the circuit including the internal resistance $1 \, \Omega$ is $R_{total} = R_p + 1 = 4 + 1 = 5 \, \Omega$.The current from the cell is $I = \frac{E}{R_{total}} = \frac{60}{5} = 12 \, A$.

The current from the cell for the given circuit is ................ $A$.

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