Capacitor's Voltage Drop And Discharge Current
Strings (SiPjAjk) = S7P3A32 Base Sequence = 12735 String Sequence = 12735 - 3 - 32
A capacitor C is connected in series with a resistor, R (fig.2.25).
Capacitor's Capacitance = 1 μF; initial charge = 10-4 C (Coulomb)'
Assume constant discharge current during 0 < t <1 ms.
Determine capacitor's voltage drop at t = 1 ms, for:
(a) R = 1 MΩ
(b) R = 100 kΩ
(c) R = 10 kΩ
(d) Determine actual discharge current.
(a) S7P3A32 (Force - Push).
Pj Problem of Interest is of type force. Problems of voltage, energy, and work in an electric circuit are force problems.
The relationship between charge (q), capacitance (C) and voltage (V) is as follows:
q = CV -------(1)
So, dq/dt = C(dv/dt)
So, i = C(dv/dt) ------- (2). Where i = current.
Discharge current has negative polarity
So, discharge current, i = -C(dv/dt)
So, v = iR = -RC(dv/dt)
So, dt = -RC(1/v)dv
So, t = -RCln(V/V0)
So, V = V0e-t/RC -------(3)
V is the voltage at time t secs during discharge; V0 is initial voltage at tim t = 0; R is the resistance; C is the capacitance.
The product RC is called the time constant.
Capacitor's voltage drop after t secs of discharge = V0 - V = V0[1 - V0e-t/(RC)]
Initial voltage V0 = q/C (from eqn (1))
So, V0 = 10-4/10-6 = 100 V.
So, capacitor's voltage drop after t = 1 ms of discharge:
(a) Time constant = t/(RC) = 10-3/(106 x 10-6) = 10-3
So, voltage drop = 100[1 - e-t/(RC)] = 100[1 - e-10-3] = 0.1 V
(b) Time constant = t/(RC) = 10-3/(105 x 10-6) = 10-2
So, voltage drop = 100[1 - e-t/(RC)] = 100[1 - e-10-2] = 1 V
(c) Time constant = t/(RC) = 10-3/(104 x 10-6) = 10-1
So, voltage drop = 100[1 - e-t/(RC)] = 100[1 - e-10-1] = 9.5 V
(d) Actual discharge current, i = (100/R)e-t/RC.
It is evident from problem (a), (b) and (c) that the capacitor's voltage drop increases as the resistance of the resistor decreases.
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