Question

In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

Hint:

Calculate the equivalent resistance of the circuit. Then use Ohm's law and current division to find the current through the ammeter.

Answer 1

Correct Answer: 5

$R_{eq} = R_1 + R_2$

$R_{eq} = 150.4 + \frac{240 \times 10}{250}$

$R_{eq} = 150.4 + 9.6 = 160 \Omega$

$I_1 = \frac{IR_2}{240}$

$I_1 = \frac{I \times 9.6}{240}$

$I = \frac{20}{160}$

$I_1 = \frac{20}{160} \times \frac{9.6}{240} = \frac{1}{200} = 5 \times 10^{-3} A = 5 mA$