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Current Electricity

Question

Wired Faculty App

A voltmeter of resistance 20000 Ω reads 5 volt. To make it read 20 volt, the extra resistance required is

40000 Ω in parallel

60000 Ω in parallel

60000 Ω in series

40000 Ω in series

Solution

C.

60000 Ω in series

$\frac{5}{20} = \frac{G}{G + R}$

$\frac{1}{4} = \frac{G}{G + R}$

⇒ R = 3 G

= 3 × 20000

⇒ R= 6000

Extra resistance = 60000 Ω in series

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