Question

What is the energy (in J) stored in the 23.0 µF capacitor of a heart defibrillator charged to 7.40 ✕ 103 V?

Answer 1

629.7 J = 0.63 kJ.

Step-by-step explanation:

Charge and voltage are related to the capacitance C of a capacitor by Q = CV, and so the expression for Ecap can be written mathematically into three equivalent expressions:

Ecap = QV^2

= 1/2 * (C*V^2)

= 1/2 * Q^2/C

where,

Q = charge

V = voltage on a capacitor C.

The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

Given:

C = 23.0 µF

= 2.3 x 10^-5 F

V = 7.40 ✕ 10^3 V

Ecap = 1/2 * (C*V^2)

= 1/2 * (2.3 x 10^-5) * (7.40 ✕ 10^3)^2

= 629.7 J

= 0.63 kJ.

Answer 2

629.74 J

Step-by-step explanation:

Energy: This can be defined as the ability or the capacity to do work.

The S.I unit of energy is Joules (J).

The formula for the energy stored in a capacitor is given as

E = 1/2CV² .............................. Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the heart defibrillator capacitor, V = voltage of the heart defibrillator .

Given: C = 23 µF = 23×10⁻⁶ F, V = 7.4×10³ V.

Substitute into equation 1

E = 1/2(23×10⁻⁶)(7.4×10³)²

E = 629.74 Joules

Hence the energy stored in the the capacitor of a heart defibrillator = 629.74 J