Question

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`

Answer 1

Correct Answer - 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`