`A 40 muF` capacitor in a defibrillator is charged to `3000 V`. The energy stored in the capacitor is sent through the patient during a pulse of duration `2 ms`. The power delivered to the patient is

Options

A. 45 kW
B. 90 kW
C. 180 kW
D. 360 kW

Answer

To solve the problem, we need to find the power delivered to a patient by a defibrillator, which involves a capacitor. Here's a step-by-step solution:

### Step 1: Calculate the Energy Stored in the Capacitor
The energy (E) stored in a capacitor can be calculated using the formula:
\[ E = \frac{1}{2} C V^2 \]
where:
- \(C\) is the capacitance in farads,
- \(V\) is the voltage in volts.

Given:
- \(C = 40 \, \mu F = 40 \times 10^{-6} \, F\)
- \(V = 3000 \, V\)

Substituting the values into the formula:
\[ E = \frac{1}{2} \times (40 \times 10^{-6}) \times (3000)^2 \]

Calculating \( (3000)^2 \):
\[ (3000)^2 = 9000000 \]

Now substituting back:
\[ E = \frac{1}{2} \times (40 \times 10^{-6}) \times 9000000 \]
\[ E = 20 \times 10^{-6} \times 9000000 \]
\[ E = 180 \, J \]

### Step 2: Calculate the Power Delivered to the Patient
Power (P) is defined as the energy delivered per unit time:
\[ P = \frac{E}{t} \]
where:
- \(E\) is the energy in joules,
- \(t\) is the time in seconds.

Given:
- \(t = 2 \, ms = 2 \times 10^{-3} \, s\)

Substituting the values:
\[ P = \frac{180 \, J}{2 \times 10^{-3} \, s} \]
\[ P = \frac{180}{0.002} \]
\[ P = 90000 \, W \]
\[ P = 90 \, kW \]

### Final Answer
The power delivered to the patient is:
\[ \text{Power} = 90 \, kW \]