The current I drawn from the 5 V source will be:

The given circuit can be redrawn as

It is a balanced Wheatstone bridge and hence. No current flows in the middle resistor. So, equivalent circuit would be as shown in the figure.

10Ω and 20Ω resistance are in series, so their equivalent resistance is

∴ R' = 10Ω + 20Ω = 30Ω

Similarly, 5Ω and 10Ω are in series, so their equivalent resistance is

R'' = 15Ω

R’ and R” are in parallel, so the equivalent resistance of the circuit is

R = $\frac{15 × 30}{15+30}$ = 10Ω                 ∴  I = $\frac{V}{R}=\frac{5 V}{10Ω}$ = 0.5 A