First slide

voltmeter

Question

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

Moderate

A

0.01
B

0.02
C

0.03
D

0.05

Solution

Before connecting voltmeter, potential difference across $400 Ω$ resistance is
$V_{i} = \frac{400}{(400 + 800)} \times 6 = 2 V$
After connecting voltmeter, equivalent resistance
between A and B $\frac{400 \times 10000}{(400 + 10000)} = 384 .6 Ω$
Hence, potential difference measured by voltmeter
$V_{f} = \frac{384 .6}{(384 .6 + 800)} \times 6 = 1 .95 V$
Error in measurement $= V_{i} - V_{f} = 2 - 1 .95 = 0 .05 V$

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