Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?

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?

Grade:11

1 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

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)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free