Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
I am unable to solve questions 2 and 3.please help.hope to to get answer soon
Always use this method to solve the question similar to this:Initial charge across capacitor 100 pF,Q = CV => 100 x 10^-12 x 24 = > 24 x 10^10 C.Now when connected with other Capacitor 0f 20 pF then assume charges Q and Q-x in capicitors resecrvely.Thus,(Q-x)/C1 – x/C2 = 0we get,x = QC2/(C2 – C1)putting the values we get,x = 6 x 10^-10 C.we get Q-x = 18 x 10^-10 C.Hence,Voltage = 18 x 10^-10/100 x 10^-12 => 18 V
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -