Current Electricity Part 4

Concept:In steady state, a fully charged capacitor acts as an open circuit.Therefore, the branch containing the $10\ \mu\mathrm{F}$ capacitor and $8\ \Omega$ resistor carries no current.Explanation:Remove the capacitor branch (open circuit).The two horizontal $8\ \Omega$ resistors on the right are in parallel.Their equivalent is $R_{4,5} = \frac{8 \times 8}{8+8} = 4\ \Omega$.This $4\ \Omega$ is in series with the $4\ \Omega$ vertical resistor ($R_3$).So $R_{3,4,5} = 4 + 4 = 8\ \Omega$.This $8\ \Omega$ is in parallel with the middle vertical $8\ \Omega$ ($R_2$).Their equivalent is $R_{2,3,4,5} = \frac{8 \times 8}{8+8} = 4\ \Omega$.Now this $4\ \Omega$ is in series with the $1\ \Omega$ resistor ($R_1$).Total resistance $R_{\text{total}} = 1 + 4 = 5\ \Omega$.Total current from battery: $i = \frac{V}{R_{\text{total}}} = \frac{10}{5} = 2\ \text{A}$.This current $i$ flows through $R_1$ into the junction.At the junction, $i$ splits between the two equal $8\ \Omega$ branches.Since the resistances are equal, current divides equally: $i_2 = i_3 = \frac{i}{2}$.The ammeter is in the branch with $R_{3,4,5}$, so it reads $i_3 = \frac{2}{2} = 1\ \text{A}$.Answer:1 A (option B).