Home
Class 12
PHYSICS
LetC be the capacitance of a capacitor d...

LetC be the capacitance of a capacitor discharging through a resistor R. Suppose `t_1` is the time taken for the energy stored in the capacitor to reduce to half its initial value and `t_2` is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio `t_1//t_2` will be

A

2

B

1

C

`(1)/(2)`

D

`(1)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, for discharging of capacitor
`V=V_(0)e^(-t//RC), Q=Q_(0)e^(-t//RC)`
and `U=(1)/(2)CV^(2)=(1)/(2)CV_(0)^(2)e^((-2t)/(RC))=U_(0)e^((-2t)/(RC))`
and `U_(0)=(1)/(2)CV_(0)^(2)`
Now, `(U_(0))/(2)=U_(0)e^(-(2t_(1))/(RC))impliest_(1)=(RCl n2)/(2)`
`(Q_(0))/(4)=Q_(0)e^((-t_(2))/(RC))impliest_(2)=2RCl n2implies(t_(1))/(t_(2))=(1)/(4)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

In a decaying L-R circuit, the time after which energy stores in the inductor reduces to one fourth of its initial values is

A charged capacitor is connected with a resistor. After how many time constants, does the energy of capacitor (1/100)^th becomes of its initial value?

In a chrging cirucit, after how many time constant will the energy stored in the capacitor reach one - half of its equilibrium value?

A parallel plate capacitor of 1 muF capacity is discharging thorugh a resistor. If its energy reduces to half in one second. The value of resistance will be

A capacitor discharges through a resistance. The stored energy U_0 in one capacitive time constant falls to

The capacitor of capacitance C in the circuit shown in fully charged initially. Resistance is R.– After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is

A charged capacitor is allowed to dischare through a resistor by closing the key at the instant t=0 . At the instant t=(ln 4)mus , the reading of the ammeter falls half the initial vaslue. The resistance of the ammeter is equal to