Current Electricity Question 4
Question 4 - 2024 (01 Feb Shift 2)
In an ammeter, $5 \%$ of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be :
[We changed options. In official NTA paper no option was correct.]
(1) $\frac{G}{20}$
(2) $\frac{G}{199}$
(3) $199 G$
(4) $200 G$
Show Answer
Answer: (1)
Solution:
$I _S S=I _g G$
$\frac{95}{100} I S=\frac{5 I}{100} G$
$S=\frac{G}{19}$ $R _A=\frac{SG}{S+G}=\frac{\frac{G^{2}}{19}}{\frac{20 G}{19}}$
$R _A=\frac{G}{20}$
