Question #c823d

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Question #c823d

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Answer 1 · 2017-01-21T09:52:39

Capacitance of 1st capacitor $C_{1} = 19 μ F$ $C_1=19muF$
Capacitance of 2nd capacitor $C_{2} = 4 μ F$ $C_2=4muF$
When connected in series their equivalent total or capacitance becomes $C_{T}$ $C_T$

So $C_{T} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}} = \frac{19 \times 4}{19 + 4} = \frac{76}{23} μ F$ $C_T=(C_1xxC_2)/(C_1+C_2)=(19xx4)/(19+4)=76/23muF$

The combination is connected in series across $10 V$ $10V$ . So each capacitor will have same charge and that will be

$Q = C_{T} \times 10 = \frac{76}{23} μ F \times 10 V = \frac{760}{23} μ C$ $Q=C_Txx10=76/23muFxx10V=760/23muC$

So P.D across 1st capacitor $\frac{Q}{C_{1}} = \frac{\frac{760}{23} μ C}{19 μ F} = \frac{40}{23} V$ $Q/C_1=(760/23muC)/(19muF)=40/23V$

And P.D across 2ndcapacitor $\frac{Q}{C_{2}} = \frac{\frac{760}{23} μ C}{4 μ F} = \frac{190}{23} V$ $Q/C_2=(760/23muC)/(4muF)=190/23V$

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