Current Electricity Question 2
Question 2 - 2024 (01 Feb Shift 1)
A galvanometer has a resistance of $50 \Omega$ and it allows maximum current of $5 mA$. It can be converted into voltmeter to measure upto $100 V$ by connecting in series a resistor of resistance
(1) $5975 \Omega$
(2) $20050 \Omega$
(3) $19950 \Omega$
(4) $19500 \Omega$
Show Answer
Answer: (3)
Solution:
$R=\frac{V}{I _g}-R _g=\frac{100}{5 \times 10^{-3}}-50$
$=20000-50$
$=19950 \Omega$
