A 40uF capacitor that had been charged to 30V is discharged through a resistor.

Question

A 40uF capacitor that had been charged to 30V is discharged through a resistor. The figure (Figure 1) shows the capacitor voltage as a function of time. What is the value of the resistance? Express your answer using two significant figures.

PLEASE EXPLAIN HOW YOU GOT THE ANSWER!!!! in detail please I don't understand how to do it.

A 40uF capacitor that had been charged to 30V is discharged through a resistor. The figure (Figure 1) shows the capacitor voltage as a function of time. What is the value of the resistance? Express your answer using two significant figures. PLEASE EXPLAIN HOW YOU GOT THE ANSWER!!!! in detail please I don't understand how to do it.

Explanation / Answer

This is a series RC discharge problem. The equation for this is: v(t) = v0 * e^(-t/RC) from the graph, we know V0 = 30 V take t=2 ms v(2 ms) = 10 V 10 = 30 * e^(-2*10^-3/(R*40*10^-6)) solve for R to get R= 45.51 ohms

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