The power delivered by the battery shown in the figure is- (Take the forward and reverse biased resistances of the diode are zero and infinity respectively)
Show Hint
For circuits with ideal diodes:
- A forward-biased diode behaves like a short circuit (zero resistance).
- A reverse-biased diode behaves like an open circuit (infinite resistance).
Simplify the circuit accordingly before solving.
Step 1: Identifying the Conducting Path - Given that the diodes are ideal, \(D_1\) is forward biased (conducting), and \(D_2\) is reverse biased (non-conducting). - The circuit simplifies to a series circuit with a \(10\Omega\) resistor connected to a \(5V\) battery.
Step 2: Calculating the Current By Ohm’s Law: \[ I = \frac{V}{R} = \frac{5V}{10\Omega} = 0.5A \]
Step 3: Power Delivered by the Battery Power is given by: \[ P = V \times I \] \[ P = 5V \times 0.5A = 2.50W \] Thus, the correct answer is \( \mathbf{(1)} \ 2.50W \).
