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) $199 \mathrm{G}$
(4) $200 \mathrm{G}$
Show Answer
Answer: (1)
Solution:
$I_{S} S=I_{g} G$
$\frac{95}{100} I S=\frac{5 I}{100} G$
$\mathrm{S}=\frac{\mathrm{G}}{19}$ $\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}=\frac{\frac{\mathrm{G}^{2}}{19}}{\frac{20 \mathrm{G}}{19}}$
$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{G}}{20}$
