Two resistances of 400 Omega and 800 Omega are connected in series with 6, volt battery of negligible internal resistance. A voltmeter of resistance 10,000

Q: Two resistances of $400$ $\Omega$ and $800$ $\Omega$ are connected in series with $6\, volt$ battery of negligible internal resistance. A voltmeter of resistance $10,000$ $\Omega$ is used to measure the potential difference across $400$ $\Omega$. The error in the measurement of potential difference in volts approximately is

d (d) Before connecting voltmeter potential difference across $400$ $\Omega$ resistance is ${V_i} = \frac{{400}}{{(400 + 800)}} \times 6 = 2\,V$ After connecting voltmeter equivalent resistance between $A$ and $B$ $ = \frac{{400 \times 10,000}}{{(400 + 10,000)}} = 384.6\,\Omega $ Hence, potential difference measured by voltmeter ${V_f} = \frac{{384.6}}{{(384.6 + 800)}} \times 6 = 1.95\,V$ Error in measurement = ${V_i} - {V_f} = 2 - 1.95$ = $0.05\,V.$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two resistances of $400$ $\Omega$ and $800$ $\Omega$ are connected in series with $6\, volt$ battery of negligible internal resistance. A voltmeter of resistance $10,000$ $\Omega$ is used to measure the potential difference across $400$ $\Omega$. The error in the measurement of potential difference in volts approximately is

A$0.01$
B$0.02$
C$0.03$
D$0.05$

Diffcult

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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