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Two resistances of 400 $Ω$ and 800 $Ω$ are connected in series with 6 V battery of negligible internal resistance. A voltmeter of resistance 10,000 $Ω$ is used to measure the p.d. across 400 $Ω$ . The error in the measurement of p.d. in volts approximately

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0.02

0.03

0.01

0.05

answer is D.

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Detailed Solution

Theoretical PD across 400 ohm resistor is $V_{0} = \frac{400}{400 + 800} \times 6 = 2 V$

Using voltmeter PD across 400 ohm resistor is, $V = \frac{(\frac{400 \times 10000}{400 + 10000})}{(\frac{400 \times 10000}{400 + 10000}) + 800} \times 6 = 1.948 V$

Error in measurement is, $2 - 1.948 = 0.052 V$

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