Find current through ammeter (in A)

Correct Option: 0.5 A

To find the current flowing through the ammeter, we analyze the circuit in its DC steady-state condition.

Step 1: Analyze the Capacitor Branch
In a DC steady state, a capacitor acts as an open circuit (infinite resistance), meaning no current flows through it. Therefore, we can completely remove the branch containing the capacitor $C$ from our calculations.

Step 2: Simplify the Right-Hand Side of the Circuit
After removing the capacitor branch, we observe the following configuration on the right side of the circuit:
The two far-right vertical resistors, each having a resistance of $8\Omega$, are connected in parallel. Their equivalent resistance ($R_p$) is calculated as:

$R_p=\frac{8\times 8}{8+8}=4\Omega$

This parallel combination is in series with the top-right horizontal resistor of $4\Omega$. Thus, the total resistance of this entire right-hand branch ($R_{\text{right}}$) is:

$R_{\text{right}}=4\Omega +4\Omega =8\Omega$

Step 3: Find the Total Equivalent Resistance of the Circuit
Now, looking at the node just after the $6\Omega$ resistor, the circuit splits into two parallel paths connected to the negative rail:
1. The branch containing the $8\Omega$ resistor and the ammeter (assuming an ideal ammeter with $0\Omega$ resistance).
2. The right-hand branch with an equivalent resistance of $R_{\text{right}}=8\Omega$.

The equivalent resistance of these two parallel branches ($R_{\text{parallel}}$) is:

$R_{\text{parallel}}=\frac{8\times 8}{8+8}=4\Omega$

This parallel combination is connected in series with the $6\Omega$ resistor at the positive terminal of the battery. Therefore, the total equivalent resistance of the entire circuit ($R_{\text{total}}$) is:

$R_{\text{total}}=6\Omega +4\Omega =10\Omega$

Step 4: Calculate the Currents
Using Ohm's law, the total current ($I_{\text{total}}$) supplied by the $10\text{V}$ battery is:

$I_{\text{total}}=\frac{10\text{V}}{10\Omega }=1\text{A}$

Since the current splits at the node after the $6\Omega$ resistor into two parallel branches of equal resistance ($8\Omega$ each), the current divides equally between them. Thus, the current flowing through the branch containing the ammeter ($I_{\text{ammeter}}$) is:

$I_{\text{ammeter}}=\frac{1\text{A}}{2}=0.5\text{A}$