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A galvanometer having internal resistance 10 \Omega requires 0.01 for a full-seale deflection. To convert this galvanometer to a voltmeter of full -scale deflection at 120V, we need to connect a resistance of

Option: 1

11990 \Omega in series


Option: 2

11990 \Omega in parallel


Option: 3

12010 \Omega  in series


Option: 4

12010 \Omega  in Parallel


Answers (1)

best_answer

The internal resistance of the galvanometer (G) is 10 \Omega. A high resistance R is to be added in series with galvanometer. In this case, if the galvanometer is connected between the two points A and B then the galvanometer will become a voltmeter of voltage range from 0 to \left(V_A-V_B\right)

According to Fig. 1.127, V_A-V_B=I_G(G+R)

\therefore R =V_A-V_B-G \\

= \frac{120}{0.01}-10 \\

= 12000-10=11990 \Omega

 

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jitender.kumar

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