The current passing through the battery in the given circuit is:

The correct answer is 0.5 A.

To find the current passing through the battery in the given circuit, we can analyze the resistor network step-by-step by identifying the nodes and simplifying the circuit connections.

Step 1: Identify the Nodes and Simplify the Circuit
By looking at the circuit diagram:
- The wire segment between points A and F has no components, which means node A and node F are at the same electrical potential (Node $A=$ Node $F$).
- Similarly, the wire segment between points C and D has no components, meaning node C and node D are at the same electrical potential (Node $C=$ Node $D$).
Now we can identify the resistors connected between the main nodes (A, B, D, and E):
- Between nodes A and B: $R_{\mathrm{AB}}=5\Omega$
- Between nodes B and D (since C = D): $R_{\mathrm{BD}}=2.5\Omega$
- Between nodes A and E (since F = A): $R_{\mathrm{AE}}=3\Omega$
- Between nodes E and D: $R_{\mathrm{ED}}=1.5\mathrm{��}$
- Between nodes A and D: $R_{\mathrm{AD}}=6\Omega$

Step 2: Analyze the Balanced Wheatstone Bridge
The network formed by the nodes A, B, D, and E with terminals at B and E represents a Wheatstone bridge where:
- The ratio of the left arms is:
$\frac{R_{\mathrm{AB}}}{R_{\mathrm{AE}}}=\frac{5}{3}$
- The ratio of the right arms is:
$\frac{R_{\mathrm{BD}}}{R_{\mathrm{ED}}}=\frac{2.5}{1.5}=\frac{5}{3}$
Since the ratio of the arms is equal:
$\frac{R_{\mathrm{AB}}}{R_{\mathrm{AE}}}=\frac{R_{\mathrm{BD}}}{R_{\mathrm{ED}}}$
The Wheatstone bridge is balanced. Consequently, no current flows through the central diagonal resistor $R_{\mathrm{AD}}=6\Omega$, and it can be removed from our calculations.

Step 3: Calculate the Equivalent Resistance of the Bridge Network ($R_{\mathrm{BE}}$)
With the $6\Omega$ resistor removed, the circuit simplifies to two parallel branches connected between terminals B and E:
1. Upper path (B → A → E):
$R_{\mathrm{BAE}}=5\Omega +3\Omega =8\Omega$
2. Lower path (B → D → E):
$R_{\mathrm{BDE}}=2.5\Omega +1.5\Omega =4\Omega$
The equivalent resistance between nodes B and E is:
$R_{\mathrm{BE}}=\frac{R_{\mathrm{BAE}}\times R_{\mathrm{BDE}}}{R_{\mathrm{BAE}}+R_{\mathrm{BDE}}}=\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}\Omega$

Step 4: Find the Total Equivalent Resistance of the Circuit
The bridge network ($R_{\mathrm{BE}}$) is connected in series with the rest of the loop containing:
- The $1.5\Omega$ resistor at the top
- The $5.5\Omega$ resistor on the right
- The $\frac{1}{3}\Omega$ resistor at the bottom
The total resistance $R_{\mathrm{total}}$ is:
$R_{\mathrm{total}}=R_{\mathrm{BE}}+1.5\Omega +5.5\Omega +\frac{1}{3}\Omega$
$R_{\mathrm{total}}=\frac{8}{3}+7+\frac{1}{3}$
$R_{\mathrm{total}}=\left(\frac{8}{3}+\frac{1}{3}\right)+7=\frac{9}{3}+7=3+7=10\Omega$

Step 5: Calculate the Total Current
Using Ohm's Law, the current $I$ passing through the battery is:
$I=\frac{V}{R_{\mathrm{total}}}=\frac{5V}{10\Omega }=0.5A$