A 1.0 μF capacitor is charged until it acquires a potential difference of 700.0 V across its plates,then the emf source is removed.If the capacitor is then discharged through a 200.0 kΩ resistance of a circuit,what is the voltage drop across the capacitor 7.0 ms later?
A)680 V
B)720 V
C)24 V
D)-25 V

Question

Quizplus · Accepted Answer

The voltage across a capacitor in a discharging circuit is given by:```V(t) = V0 * e^(-t/RC)```where:* V(t) is the voltage across the capacitor at time t* V0 is the initial voltage across the capacitor* R is the resistance in the circuit* C is the capacitance of the capacitorIn this problem, we have:* V0 = 700.0 V* R = 200.0 kΩ = 200,000 Ω* C = 1.0 μF = 10^-6 F* t = 7.0 ms = 0.007 sPlugging these values into the equation, we get:```V(0.007 s) = 700.0 V * e^(-0.007 s / (200,000 Ω * 10^-6 F))``````V(0.007 s) = 680.0 V```Therefore, the voltage drop across the capacitor 7.0 ms later is 680 V.

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