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Current Electricity question

2022 · 24 Jun · Shift 1 · Q46

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Current Electricity
2022 · 24 Jun · Shift 1 · Q46

Current Electricity question

2022 · 24 Jun · Shift 1 · Q46

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20

\Omega

. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

A
2.5 $\Omega$
B
4 $\Omega$
C
5 $\Omega$
D
10 $\Omega$

View written solutionFree

Correct answer: C

Find the equivalent external resistance

Two resistors of $20\,\Omega$ each are connected in parallel.

R_{\text{ext}}=\frac{20\times 20}{20+20}=\frac{400}{40}=10\,\Omega

Find the equivalent emf of the cells

Two identical cells, each of emf $1.5\,\text{V}$ , are connected in parallel. For identical cells in parallel, the equivalent emf remains the same:

E_{\text{eq}}=1.5\,\text{V}

If the internal resistance of each cell is $r$ , then the equivalent internal resistance in parallel is:

r_{\text{eq}}=\frac{r}{2}

Use the voltmeter reading

The voltmeter reads the terminal voltage across the external resistance, so

V=1.2\,\text{V}

Current through the external circuit:

I=\frac{V}{R_{\text{ext}}}=\frac{1.2}{10}=0.12\,\text{A}

Relate emf, terminal voltage, and internal drop

For the equivalent cell combination,

E = V + I r_{\text{eq}}

So,

1.5 = 1.2 + 0.12\left(\frac{r}{2}\right)

0.3 = 0.12\cdot \frac{r}{2}

0.3 = 0.06r

r=\frac{0.3}{0.06}=5\,\Omega

Check the options

A: $2.5\,\Omega$ ❌
B: $4\,\Omega$ ❌
C: $5\,\Omega$ ✅
D: $10\,\Omega$ ❌

Therefore, the internal resistance of each cell is:

\boxed{5\,\Omega}

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