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Current Electricity question

2020 · 3 Sep · Shift 2 · Q56

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JEE Main
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Current Electricity
2020 · 3 Sep · Shift 2 · Q56

Current Electricity question

2020 · 3 Sep · Shift 2 · Q56

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1

Two resistors 400

\Omega

and 800

\Omega

are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k

\Omega

across 400

\Omega

resistor is close to :

A
2.05 V
B
1.95 V
C
2 V
D
1.8 V

View written solutionFree

Correct answer: B

Given data

Resistors in series: $R_1 = 400\,\Omega$ , $R_2 = 800\,\Omega$
Battery voltage: $V = 6\,\text{V}$
Voltmeter resistance: $R_v = 10\,\text{k}\Omega = 10000\,\Omega$

We need the voltmeter reading when it is connected across the $400\,\Omega$ resistor.

Effect of connecting the voltmeter

Since the voltmeter is connected across the $400\,\Omega$ resistor, it is in parallel with it.

So the equivalent resistance of this parallel combination is

R_p = \frac{400 \times 10000}{400 + 10000}

R_p = \frac{4000000}{10400} \approx 384.615\,\Omega

Total resistance of the circuit

Now this parallel combination is in series with the $800\,\Omega$ resistor:

R_{\text{total}} = R_p + 800 = 384.615 + 800 = 1184.615\,\Omega

Total current from the battery

I = \frac{V}{R_{\text{total}}} = \frac{6}{1184.615}

I \approx 5.065 \times 10^{-3}\,\text{A}

Voltage across the parallel combination

The voltmeter reads the voltage across the parallel branch, i.e. across the $400\,\Omega$ resistor:

V_{400} = I R_p

V_{400} = \left(5.065 \times 10^{-3}\right)(384.615)

V_{400} \approx 1.95\,\text{V}

Check options

A: $2.05\,\text{V}$
B: $1.95\,\text{V}$
C: $2\,\text{V}$
D: $1.8\,\text{V}$

So the closest value is:

\boxed{1.95\,\text{V}}

Hence, Option B is correct.

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