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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` |
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Answer» 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` |
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