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JAMB - Physics (2005 - No. 42)

The figure above shows three capacitors, 2μF, 3μF and 6μF connected in series. If the p.d across the system is 12V, the p.d, across the 6μF capacitor is
4V
12V
2V
6V

Explication

The effective capacitance in the circuit is given by:
1/C = 1/2 + 1/3 + 1/6 = 6/6
therefore C = 1μF or
1.0 x 10-6F.
Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV
= 1.0 x 10-6 x 12
= 12.0 x 10-6 Coulomb.
∴ P.d across the 6μF,
V = Q/C = 12.0 x 10-6
6.0 x 10-6F

= 2.0 volts

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