A) 0.01
B) 0.02
C) 0.03
D) 0.05
Correct Answer: D
Solution :
Before connecting voltmeter potential difference across 400W resistance is \[{{V}_{i}}=\frac{400}{(400+800)}\times 6=2V\] 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 \[{{V}_{f}}=\frac{384.6}{(384.6+800)}\times 6=1.95V\] Error in measurement = \[{{V}_{i}}-{{V}_{f}}=2-1.95\] = 0.05V.You need to login to perform this action.
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