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Two capacitors are 2and3 uF and are joined in series. The outer plate of the first capacitor is at 1000V and the outer plate of the second capacitor is earthed. Find out the potential and charge of the inner plate of each capacitor |
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Answer» ● Answer -Q1 = 1200 μC, V1 = 600 VQ2 = 1200 μC, V2 = 400 V◆ Explaination -When two CAPACITORS ATR connected in series, EQUIVALENT capacitance is given by -C = C1.C2 / (C1+C2)C = 2 × 3 / (2 + 3)C = 6 / 5C = 1.2 μFAs capacitors are connected in series, charge will be same.Q = Q1 = Q2Q = C.VQ = 1.2 × 1000Q = 1200 μCPotential of INNER plate of 2 μF capacitor -V1 = Q / C1V1 = 1200 / 2V1 = 600 VPotential of inner plate of 3 μF capacitor -V2 = Q / C2V2 = 1200 / 3V2 = 400 Vmark me as brainlist... |
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