A 40 μ F 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

Question

A 40 μ F 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 (A) 45 kW (B) 90 kW (C) 180 kW (D) 360 kW

Tardigrade Admin · Accepted Answer

A capacitor is a device that stores energy in the electric field created between a pair of conductors on which equal but opposite electric charges have been placed.
The energy stored in a capacitor

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

Given,

C = 40 μ F = 40 \times 1 0^{- 6} F, V = 3000 V

∴ E = \frac{1}{2} \times 40 \times 1 0^{- 6} \times (3000)^{2}

= 180 J

Also

1 W = 1 J / s

∴ 2 m s = 2 \times 1 0^{- 3} s

Hence, power

= \frac{180 J}{2 \times 1 0 ^{- 3} s}

= 90 \times 1 0^{3} W = 90 kW

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

45 kW

90 kW

180 kW

360 kW

Q. A $40 μ F$ 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 $2 m s$ . The power delivered to the patient is