1.

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. `1M Omega`B. `1 Omega`C. `2 Omega`D. `2M Omega`

Answer» Correct Answer - C
`I = I_(0)e^(-t//tau)`
`2=e^(t//tau) rArr ln2 = (t)/(tau) rArr tau = (ln4)/(ln2) rArr tau = 2muS`
`RC = 2xx10^(-6) rArr (2+R_(A)) xx 0.5 xx 10^(-6) =2xx10^(-6)`
`2+R_(A) = 4 rArr R_(A) = 2 Omega`


Discussion

No Comment Found

Related InterviewSolutions