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4muFcapacitor and a resistance 2.5MOmegaare in series with 12V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given In ( 2 ) = 0.693] |
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Answer» Solution :CHARGING current `i=(V_0)/(R ) e^((t)/(RC))` `therefore` Potential difference across R is `V_R = IR = V_0e^(-(t)/(RC))` `therefore` Potential difference across .C. is `V_C = V_0 - V_R` `= V_0 - V_0e^((t)/(RC)) = V_0 (1-e^((t)/(RC)) )` but given `V_C = 3V_R` , we get `1-e^(-t//RC) = 3e^(-t//RC) or 1 = 4E^(-t//RC)` `e^((t)/(RC)) = 4 rArr (t)/(RC) = ln 4 rArr t= 2RC ln 2` `t = 2.5 xx 10^6 xx 4 xx 10^(-6) xx 2 xx 0.693` or t = 13.86 sec |
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