Question

A 44.4 ohm resistor is in series with a 5.86 volt power supply and and a capacitor. 15.8 ms after the power supply is turned on, the capacitor reaches 85.1 of its maximum charge. What is the capacitance?

Answer 1

To determine the capacitor's capacitance, the charge equation at a certain time is used with given values. After calculations, the capacitance is found to be approximately 187 µF.

Step-by-step explanation:

To find the capacitance of the capacitor, we can use the formula that relates the charge on a capacitor at a certain time to its maximum charge. This is Q(t) = Qmax(1 - e-t/RC), where Q(t) is the charge at time t, Qmax is the maximum charge, R is the resistance, C is the capacitance, and t is the time.

The problem states that the capacitor reaches 85.1% of its maximum charge after 15.8 ms, so we can set up our equation like this: 0.851Qmax = Qmax(1 - e-t/RC). We know the time t = 15.8 ms and the resistance R = 44.4 ohms.

Now we will solve for RC:

0.851 = 1 - e-t/RC
=> e-t/RC = 1 - 0.851
=> e-t/RC = 0.149

Now let's solve for t/RC:

-t/RC = ln(0.149)
=> RC = -t/ln(0.149)

Substitute the known values:

RC = -(15.8 x 10-3 s)/(ln(0.149))
=> RC ≈ 15.8 x 10-3 s / 1.9042
=> RC ≈ 8.30 x 10-3 s

Now we can solve for C:

C = RC / R
=> C ≈ (8.30 x 10-3 s) / (44.4 ohms)
=> C ≈ 1.87 x 10-4 F or 187 µF

Therefore, the capacitance of the capacitor is approximately 187 µF.