In the given circuit, what is the reading of | vedclass

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(A) The total current $I = 2 \, A$ splits into two parallel branches at point $X$. Since the resistance of the upper branch is $4 \, \Omega + 16 \, \O

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The total current $I = 2 \, A$ splits into two parallel branches at point $X$. Since the resistance of the upper branch is $4 \, \Omega + 16 \, \Omega = 20 \, \Omega$ and the lower branch is $16 \, \Omega + 4 \, \Omega = 20 \, \Omega$, the current divides equally, i.e., $1 \, A$ in each branch.Let the potential at $X$ be $V_X = 0 \, V$ for reference.The potential at point $Y$ is $V_Y = V_X - I_1 R_1 = 0 - (1 \, A 	imes 4 \, \Omega) = -4 \, V$.The potential at point $Z$ is $V_Z = V_X - I_2 R_2 = 0 - (1 \, A 	imes 16 \, \Omega) = -16 \, V$.The voltmeter reading is the potential difference between $Y$ and $Z$, which is $|V_Y - V_Z| = |-4 \, V - (-16 \, V)| = |-4 \, V + 16 \, V| = 12 \, V$.

In the given circuit, what is the reading of the voltmeter in $V$?

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