A capacitor of capacitance c connected to dc voltage a charged to the potential v how the displacement current and conduction current will be changed if
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Look for Attachment for charge Distribution in Capacitor .
This will certainly Boost your thinking, (•_^)
Final Answer : 2.25
Steps:
1) Initially, Capacitor will be charged to chage q(i) = CV on one side of plate capacitor .
After that, on changing the polarity of Battery and of EMF 2V ,Final charge on same side of plate capacitor q(f) = -C (2V) .
2) Hence, magnitude of charge transferred by Battery of 2V is
| CV - (-2CV)| = 3CV
Now,
Work done by battery = Charge transferred * EMF
= (3CV) *(2V) = 6CV^2
3) Change in Electrostatic Energy of Capacitor,
del(U) =
4)
W(B) = del(U) + Heat
Don't fear , this just Energy Conservation •_^
5) Now,
Ratio of Heat Generated to Final Energy of Capacitor = Heat / Final Energy of Capacitor
Hence, Required Ratio is 2.25
This will certainly Boost your thinking, (•_^)
Final Answer : 2.25
Steps:
1) Initially, Capacitor will be charged to chage q(i) = CV on one side of plate capacitor .
After that, on changing the polarity of Battery and of EMF 2V ,Final charge on same side of plate capacitor q(f) = -C (2V) .
2) Hence, magnitude of charge transferred by Battery of 2V is
| CV - (-2CV)| = 3CV
Now,
Work done by battery = Charge transferred * EMF
= (3CV) *(2V) = 6CV^2
3) Change in Electrostatic Energy of Capacitor,
del(U) =
4)
W(B) = del(U) + Heat
Don't fear , this just Energy Conservation •_^
5) Now,
Ratio of Heat Generated to Final Energy of Capacitor = Heat / Final Energy of Capacitor
Hence, Required Ratio is 2.25
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