| A resistance R is measured using a voltmeter and an ammeter as shown below. |
 |
| Ammeter shows 2A and voltmeter shows 120V. |
| Internal resistance of voltmeter is\[3000\Omega \]. Error in measurement of R compared to the reading taken with an ideal voltmeter will be |
A) \[3.2\Omega \]
B) \[4.2\Omega \]
C) \[1.2\Omega \]
D) \[0.2\Omega \]
Correct Answer: C
Solution :
| Let I = ammeter reading and |
| V= voltmeter reading. |
| \[V=\frac{{{R}_{V}}.R}{{{R}_{V}}+R}.I\] |
| where, \[{{R}_{V}}\]is the voltmeter resistance. |
| or \[R=\frac{V{{R}_{V}}}{I{{R}_{V}}-V}\] |
| \[\Rightarrow R=\frac{120\times 3000}{2\times 3000-120}=61.2\Omega \] |
| when\[{{R}_{V}}=\infty \], i.e. voltmeter is ideal, |
| So, error in \[R=61.2-60=1.2\Omega \] |
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