A resistance of 6*10^6 ohm and a capacitor of 2F areconnected in a series RC cir

Question

A resistance of 6*10^6 ohm and a capacitor of 2F areconnected in a series RC circuit across a 10-V battery. a) What is the time constant of the circuit? b) If the capacitor were initially uncharged, what is thecapacitor voltage 25 seconds after the battery is connected? c) If the capacitor were discharged from a fully chargedcondition at 10 V by disconnecting the battery, how long does ittake the capacitor voltage to drop to 2V? A resistance of 6*10^6 ohm and a capacitor of 2F areconnected in a series RC circuit across a 10-V battery. a) What is the time constant of the circuit? b) If the capacitor were initially uncharged, what is thecapacitor voltage 25 seconds after the battery is connected? c) If the capacitor were discharged from a fully chargedcondition at 10 V by disconnecting the battery, how long does ittake the capacitor voltage to drop to 2V?

Explanation / Answer

The resistance in the series RC circuit is R = 6 *106 ohm The capacitance in the series RC circuit is C = 2 F = 2* 10-6 F The battery in the circuit is Vo = 10-V a)The time constant of the circuit is = R * C or = 6 * 106 * 2 * 10-6 = 12s b)The voltage across the capacitor is V = Vo* e(-t/RC) Here,t = 25 s or V = 10 * e(-25/6 * 10^6 * 2 * 10^-6) or V = (10/2.083e) = (10/2.083 * 2.71) = 1.77 V c)The capacitor voltage drops to 2 V,therefore,we get V = Vo* e(-t/RC) V = Vo* e(-t/RC) Here,V = 2 V and Vo= 10 V or 2 = 10 * e(-t/RC) or e(-t/RC)= (1/5) or (-t/RC) = ln(1/5) or (t/RC) = -ln(1/5) = ln(5) or t = RC * ln(5) Substituting the values in the above equation,we get t = 6 * 106 * 2 * 10-6 * ln(5) or t = 19.3 s Therefore,it takes 19.3 s for the capacitor voltage to drop to2 V. or e(-t/RC)= (1/5) or (-t/RC) = ln(1/5) or (t/RC) = -ln(1/5) = ln(5) or t = RC * ln(5) Substituting the values in the above equation,we get t = 6 * 106 * 2 * 10-6 * ln(5) or t = 19.3 s Therefore,it takes 19.3 s for the capacitor voltage to drop to2 V.

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