In the circuit shown below,the reading of the | vedclass

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(A) The total current of $2 \, A$ enters the junction $C$ and splits into two parallel branches. Since the resistances in the upper branch $(4 \, \Ome

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The total current of $2 \, A$ enters the junction $C$ and splits into two parallel branches. Since the resistances in the upper branch $(4 \, \Omega + 16 \, \Omega = 20 \, \Omega)$ and the lower branch $(16 \, \Omega + 4 \, \Omega = 20 \, \Omega)$ are equal,the current divides equally,i.e.,$1 \, A$ in each branch.Let the potential at point $C$ be $V_C$. The potential at point $A$ is $V_A = V_C - (1 \, A 	imes 4 \, \Omega) = V_C - 4 \, V$.The potential at point $B$ is $V_B = V_C - (1 \, A 	imes 16 \, \Omega) = V_C - 16 \, V$.The voltmeter $V$ measures the potential difference between points $A$ and $B$,which is $|V_A - V_B|$.$V_A - V_B = (V_C - 4) - (V_C - 16) = -4 + 16 = 12 \, V$.Therefore,the reading of the voltmeter is $12 \, V$.

In the circuit shown below,the reading of the voltmeter $V$ is ........... (in $, V$)

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