Since the circuit diagram is missing, we can deduce its structure from the standard AIPMT 2012 question. The circuit consists of a $5 \text{ V}$ battery connected in parallel to two branches. The first branch has an ideal diode $D_1$ in series with a $10 \text{ } \Omega$ resistor, and the second branch has an ideal diode $D_2$ in series with a $20 \text{ } \Omega$ resistor. Diode $D_1$ is forward-biased and conducts, offering zero resistance. Diode $D_2$ is reverse-biased and acts as an open circuit, so no current flows through the $20 \text{ } \Omega$ resistor. The total current supplied by the battery flows only through the $10 \text{ } \Omega$ resistor. According to Ohm's law, $I = \frac{V}{R} = \frac{5 \text{ V}}{10 \text{ } \Omega} = 0.5 \text{ A}$.