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Dear student,
Circuits with resistors and batteries have time-independent solutions: the current doesn't change as time goes by. Adding one or more capacitors changes this. The solution is then time-dependent: the current is a function of time.
Consider a series RC circuit with a battery, resistor, and capacitor in series. The capacitor is initially uncharged, but starts to charge when the switch is closed. Initially the potential difference across the resistor is the battery emf, but that steadily drops (as does the current) as the potential difference across the capacitor increases.
Applying Kirchoff's loop rule:
e - IR - Q/C = 0
As Q increases I decreases, but Q changes because there is a current I. As the current decreases Q changes more slowly.
I = dQ/dt, so the equation can be written:
e - R (dQ/dt) - Q/C = 0
This is a differential equation that can be solved for Q as a function of time. The solution (derived in the text) is:
Q(t) = Qo [ 1 - e-t/t ]
where Qo = C e and the time constant t = RC.
Differentiating this expression to get the current as a function of time gives:
I(t) = (Qo/RC) e-t/t = Io e-t/t
sir,if the circuit consists of resister and capacitor in parallel connected to the battery of emf ,now current coming from battery splits andgoes to resistor and capacitor or not ???
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