Q 12-04-081JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
An ammeter A consists of a $240\ \Omega$ coil connected in parallel to a $10\ \Omega$ shunt. It is connected in series with a $140.4\ \Omega$ resistor across a $24\ \text{V}$ battery. The reading of the ammeter is ______ mA.
Numerical value type. Enter your answer.
Answer: 160
Ammeter resistance: $\dfrac{240\times10}{250} = 9.6\ \Omega$.
$$I = \frac{24}{140.4 + 9.6} = \frac{24}{150} = 0.16\ \text{A} = 160\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics