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(C) In the given circuit, both diodes are connected in parallel and are forward biased because their p-sides are connected to the positive terminal of

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$0.5$

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(C) In the given circuit, both diodes are connected in parallel and are forward biased because their p-sides are connected to the positive terminal of the battery.Since the diodes are ideal (assuming no threshold voltage unless specified), the resistance of each branch is $20 \Omega$.The equivalent resistance $R_{	ext{eq}}$ of two $20 \Omega$ resistors in parallel is given by:$\frac{1}{R_{	ext{eq}}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$Therefore, $R_{	ext{eq}} = 10 \Omega$.The total current $i$ through the battery is given by Ohm's law:$i = \frac{V}{R_{	ext{eq}}} = \frac{5 	ext{V}}{10 \Omega} = 0.5 	ext{A}$.

What is the current through the battery in the circuit shown below (in $\text{A}$)?

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