Problem: A 1.70-μF capacitor is charging through a 10.0-Ω resistor using a 10.0-V battery.What will be the current when the capacitor has acquired 1/4 of its maximum charge? Will it be 1/4 of the maximum current?

FREE Expert Solution

Since the capacitor is charging:

Q=Q0(1-e-tRC)

New current:

I=I0e-tRC

In the given case, Q = Q0/4

Substituting:

Q04=Q0(1-e-tRC)14=1-e-tRCe-tRC=1-14e-tRC=34-tRC=ln(34)-tRC=-0.28768t=(-0.28768)(10.0)(1.70×10-6)

t = 4.89056 × 10-6s

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Problem Details

A 1.70-μF capacitor is charging through a 10.0-Ω resistor using a 10.0-V battery.

What will be the current when the capacitor has acquired 1/4 of its maximum charge? Will it be 1/4 of the maximum current?

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