A galvanometer of 25 ohms and having full scale deflection for a current of 10 mA is changed into voltmeter of range 100 volts by connecting a resistance R in series with the galvanometer. The resistance R in ohms is:

Question

A galvanometer of 25 ohms and having full scale deflection for a current of 10 mA is changed into voltmeter of range 100 volts by connecting a resistance R in series with the galvanometer. The resistance R in ohms is: (A) 10000 (B) 975 (C) 10025 (D) 9975

Tardigrade Admin · Accepted Answer

Using formula for resistance R R=(V/ig)-G Here, G=25 Ω , ig=10× 10-3A, V=100 V On putting the value in above formula, we get =(100/10× 10-3)-25 =1000-25 =9975 ohm

Q. A galvanometer of 25 ohms and having full scale deflection for a current of 10 mA is changed into voltmeter of range 100 volts by connecting a resistance R in series with the galvanometer. The resistance R in ohms is:

10000

975

10025

9975

Using formula for resistance R $R = \frac{V}{i _{g}} - G$ Here, $G = 25 Ω,$ $i_{g} = 10 \times 10^{- 3} A,$ $V = 100 V$ On putting the value in above formula, we get $= \frac{100}{10 \times 10 ^{- 3}} - 25$ $= 1000 - 25$ $= 9975 o hm$

Q. A galvanometer of 25 ohms and having full scale deflection for a current of 10 mA is changed into voltmeter of range 100 volts by connecting a resistance R in series with the galvanometer. The resistance R in ohms is:

10000

975

10025

9975

Using formula for resistance R R=ig​V​−G Here, G=25Ω, ig​=10×10−3A, V=100V On putting the value in above formula, we get =10×10−3100​−25 =1000−25 =9975ohm

Using formula for resistance R $R = \frac{V}{i _{g}} - G$ Here, $G = 25 Ω,$ $i_{g} = 10 \times 10^{- 3} A,$ $V = 100 V$ On putting the value in above formula, we get $= \frac{100}{10 \times 10 ^{- 3}} - 25$ $= 1000 - 25$ $= 9975 o hm$