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.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Current Electricity

In the figure shown below, a resistance of

150.4Ω

is connected in series to an ammeter A of resistance

240Ω

. A shunt resistance of

10Ω

is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

Answer: 5

The ammeter (

240 Ω

) and the shunt (

10 Ω

) are in parallel:

R_{p} = \frac{240 \times 10}{240 + 10} = 9.6 Ω.

With the

150.4 Ω

resistor in series the total resistance is

150.4 + 9.6 = 160 Ω

, so the

20 V

source drives

I = \frac{20}{160} = 0.125 A = 125 mA .

This current divides inversely with resistance, so the part through the ammeter is

I_{A} = 125 \times \frac{10}{240 + 10} = 5 mA .

The ammeter reads

5 mA

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer