Home
Class 12
PHYSICS
A charged capacitor is allowed to disch...

A charged capacitor is allowed to discharge 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 value. The resistance of the ammeter is equal to

A

`0`

B

`2Omega`

C

`oo`

D

`2M Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
`I = I_(0)e^(-t//tau) rArr 2=e^(t//tau) rArr ln2 = (t)/(tau)`
`tau = (ln2)/(ln2) rArr tau = 1muS rArr RC = 1xx 10^(-6)`
`(2+R_(A)) xx 0.5 xx 10^(-6) = 1xx 10^(-6)`
`2+R_(A) = 2rArr R_(A = 0`
Promotional Banner

Similar Questions

Explore conceptually related problems

A charged capacitor is allowed to discharged three a resistor 2 Omega by closing the switch S at the t=0 . At time t = 2 mus , the reading of the falls half of its initial value. The resistance of ammeter is equal to

In the circuit shown the key (K) is closed at t=0 , the current through the key at the instant t=10^-3 ln, 2 is

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

An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch S_(1) is closed the reading of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch S_(2) is also closed, then reading of ammeter becomes :

The figure-3.375 shows two circuits with a charged capacitor that is to be discharged through a resistor as shown in the figure. The ratio of initial charges on capacitors is given as q_(2)//q_(1) = 2 . If both switches are closed at time t= 0 , the charges become equal at 10^(-4) ln 2 s . Find the resistance R.

A capacitor of capacitance 0.5 mu F is discharged through a resistance. Find the value of resistance if half the charge on capacitor escapes in one minute.

An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E/2, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-

A 0.18 mu F capacitor is first charged and then discharged through a high resistance. If it takes 0.5 sec for the chage to reduce to one forth of its initial value, find the value of the resistance. Given log_(e) 4 = 1.386