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)
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 \( \mathbf{(1)} \ 2.50W \).
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given:
Match the following: