The reading of the ammeter $(A)$ in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ______ A.
Answer: (B) 1
In steady state no current flows through the capacitor, so the diagonal branch ($10\ \mu$F in series with $8\ \Omega$) carries no current.
Let $P$ be the junction after the $1\ \Omega$ resistor. From $P$ the current has two paths back to the cell:
- through the left $8\ \Omega$ directly to the bottom wire, and
- through $4\ \Omega$, then the two right-hand $8\ \Omega$ resistors in parallel ($4\ \Omega$), then through the ammeter to the bottom wire: $4 + 4 = 8\ \Omega$.
These two $8\ \Omega$ paths are in parallel: $4\ \Omega$. Total resistance $= 1 + 4 = 5\ \Omega$.
$$I = \frac{10}{5} = 2\ \text{A}$$
It divides equally between the two equal paths, so the ammeter reads $1$ A.
Solution by Sreeraj P, M.Sc Physics