AIIMS 2004 Physics Capacitors MCQ Question

To solve the problem, we first need to determine the energy stored in the capacitor, which is given by the formula:

$$U = \frac{1}{2} C V^2$$

where:

• $U$ is the energy stored in joules (J),
• $C$ is the capacitance in farads (F),
• $V$ is the voltage in volts (V).

In this case, we have:

• $C = 40 \, \mu F = 40 \times 10^{-6} \, F$,
• $V = 3000 \, V$.

Substituting these values into the energy formula:

$$U = \frac{1}{2} \times 40 \times 10^{-6} \times (3000)^2$$

Calculating $(3000)^2$:

$$(3000)^2 = 9000000$$

Now substituting back into the energy equation:

$$U = \frac{1}{2} \times 40 \times 10^{-6} \times 9000000 = 20 \times 10^{-6} \times 9000000 = 180 \, J$$

Now that we have the energy stored in the capacitor, we can calculate the power delivered to the patient. Power is defined as the energy transferred per unit time:

$$P = \frac{U}{t}$$

where:

• $P$ is the power in watts (W),
• $t$ is the time in seconds (s).

In this case, the pulse duration $t = 2 \, ms = 2 \times 10^{-3} \, s$.

Substituting the values we have:

$$P = \frac{180 \, J}{2 \times 10^{-3} \, s} = \frac{180}{0.002} = 90000 \, W = 90 \, kW$$

Thus, the power delivered to the patient is $90 \, kW$, which corresponds to option B.

Explanation of Other Options

• Option A (45 kW): This option is incorrect as it does not reflect the energy stored in the capacitor or the duration of the pulse correctly.

• Option C (180 kW): This option is also incorrect. It seems to misinterpret the energy or assumes a different time duration.

• Option D (360 kW): This option is incorrect as well; it suggests a misunderstanding of the energy-to-time ratio.

Conclusion

The correct power delivered to the patient during the pulse is $90 \, kW$, which corresponds with option B. We derived this by first calculating the energy stored in the capacitor and then determining the power based on the duration of the pulse.