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Grade: 11

                        

Two capacitors, with capacitance 4.20*10^-6 F and 5.90*10^-6 F, are connected in parallel across a 660-V supply line. The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. a) find the final charge on the 4.20*10^-6 F capacitor? b) find the final charge on the 5.90*10^-6 F capacitor? c) find the final voltage across the 4.20*10^-6 F capacitor?

9 years ago

Answers : (1)

Chetan Mandayam Nayakar
312 Points
							

Q1=C1*V,Q2=C2*V. after terminals of unlike sign are connected total charge=absolute value of Q2-Q1. It is divided between the two capacitors in such a way that the voltages across the two capacitors are equal(q1+q2=abs(Q2-Q1),q1/C1=q2/C2)

9 years ago
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