WAEC SSCE - Physics - 1995 (Objective)

The energy stored in a capacitor of capacitance 5μF is 40J. Calculate the voltage applied across its terminals?

4000V Resposta correta

200V

16V

Detalhes da Resposta

The energy stored in a capacitor can be expressed as E = 1/2 * C * V^2, where E is the energy stored in the capacitor, C is the capacitance of the capacitor, and V is the voltage applied across its terminals. From the given information, we have E = 40J and C = 5μF. Substituting these values into the equation, we get: 40J = 1/2 * 5μF * V^2 Simplifying the equation, we get: V^2 = (2*40J)/(5μF) = 16,000,000 μV^2 Taking the square root of both sides, we get: V = 4000μV = 4000V (since 1V = 1,000,000μV) Therefore, the voltage applied across the terminals of the capacitor is 4000V.






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