NEET UG 2019 — Physics Electromagnetism

2019mcqhard

In the circuits shown below, the readings of the voltmeters and the ammeters will be,

NEET 2019 Physics, Current Electricity — question figure

Official previous-year question

Held on 30 Apr 2019 · Verified 9 Jul 2026.

Options

A
${V}_{2}>{V}_{1}$ and ${i}_{1}={i}_{2}$
B
${V}_{1}={V}_{2}$ and ${i}_{1}>{i}_{2}$
C
${V}_{1}={V}_{2}$ and ${i}_{1}={i}_{2}$
D
${V}_{2}>{V}_{1}$ and ${i}_{1}>{i}_{2}$

Did you get this right?

Solution

For an ideal voltmeter, resistance is infinite $({R}_{V}=\infty )$ , so it will not draw any amount of current from the circuit and for ideal ammeter resistance is zero.

Then for the first figure ${R}_{1}=10\Omega ,{R}_{V}=\infty$ and both are in a parallel arrangement, so

$\frac{1}{{R}_{\mathrm{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{V}} \Rightarrow \frac{1}{{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{\infty } \Rightarrow {R}_{\mathrm{eq}}=10\Omega$

and current in the first circuit is, ${i}_{1}=\frac{{V}_{1}}{{R}_{\mathrm{eq}}}=\frac{10}{10}=1A$

So, for the first circuit, ammeter and voltmeter readings are ${V}_{1}=10V,{i}_{1}=1A$ , respectively.

For the second figure voltmeter and resistant of $10\Omega$ are in series and also connected parallel to another resistant of $10\Omega$ . So, equivalent resistant is

$\frac{1}{{{R}^{'}}_{\mathrm{eq}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}+{R}_{V}} \Rightarrow \frac{1}{{{R}^{'}}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10+\infty }=\frac{1}{10}+0=\frac{1}{10} \Rightarrow {{R}^{'}}_{\mathrm{eq}}=10\Omega$

$\therefore$ Current in the circuit is, ${i}_{2}=\frac{{V}_{2}}{{R}_{\mathrm{eq}}^{'}}=\frac{10}{10}=1A$ ,

$\therefore$ Ammeter and voltmeter readings for the second circuit is ${V}_{2}=10V,{i}_{2}=1A$

$\therefore$ For both the circuit is

${V}_{1}={V}_{2}=10V, {i}_{1}={i}_{2}=1A$

Your note

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