The reading in the ideal voltmeter (V) shown | vedclass

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(A) In the circuit, we have three branches connected in parallel.Branch $1$ (top): Three cells of $5 \, V$ and $0.2 \, \Omega$ each in series. Total $

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) In the circuit, we have three branches connected in parallel.Branch $1$ (top): Three cells of $5 \, V$ and $0.2 \, \Omega$ each in series. Total $EMF$ $E_1 = 5 + 5 + 5 = 15 \, V$, total internal resistance $r_1 = 0.2 + 0.2 + 0.2 = 0.6 \, \Omega$.Branch $2$ (left): One cell of $5 \, V$ and $0.2 \, \Omega$. $E_2 = 5 \, V$, $r_2 = 0.2 \, \Omega$.Branch $3$ (bottom): Three cells of $5 \, V$ and $0.2 \, \Omega$ each in series. $E_3 = 5 + 5 + 5 = 15 \, V$, $r_3 = 0.2 + 0.2 + 0.2 = 0.6 \, \Omega$.However, looking at the polarity, the top and bottom branches are connected in opposition to the left branch. Let the potential at the right junction be $V_A$ and left be $V_B$. The voltmeter measures the terminal voltage of the rightmost cell $(5 \, V, 0.2 \, \Omega)$.Actually, the circuit shows a loop where the EMFs cancel out or balance. Specifically, the rightmost branch contains a $5 \, V$ cell and a voltmeter in parallel. For an ideal voltmeter, no current flows through it. The potential difference across the $5 \, V$ cell is $V = E - Ir$. Since the circuit is balanced such that no current flows through this specific branch, $I = 0$, so $V = E = 5 \, V$.

The reading in the ideal voltmeter $(V)$ shown in the given circuit diagram is: (in $ V$)

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