V=V0(1−e−t/RC)
2=20(1−e−10C1μs)
101=(1−e−10C1×10−6)
e−10C1×10−6=109
10C1×10−6=ln(910)
C=10×ln(910)1×10−6=1.0510−7=1.051μF=105100μF=0.95μF
A capacitor is connected to a 20V battery through a resistance of 10Ω. It is found that the potential difference across the capacitor rises to 2V in 1μs. The capacitance of the capacitor is ________\mu F.
Given In (910)=0.105
Held on 1 Sept 2021 · Verified 6 Jul 2026.
0.95
9.52
1.85
0.105
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