2 What is the total energy dissipated in the resistor after
Solution
Voltage Vo = 3volt
Resistance R = 300 ohm
Capacitance C = 3 x10 -6 F
Time t= 1.5 x10 -3 s
In the dischargeing of the capacitor in RC circuit,
Charge at time t is q = Q e -t/RC
Where Q = maximum charge
qC = QC e -t/RC
We know qC = V and QC = Vo Where Vo = maximum voltage
Therefore V = Vo e -t/RC
= 3 e -(1.5 x10^-3 )/(300)(6x10^-6)
= 3 e -0.833
=3(0.4345)
=1.303 volt
(b).Total energy dissipated in the resistor P=Vo 2/R
= 3 2 / 300
= 0.03 watt
(c).
Time t= 1.1 x10 -3 s
In the dischargeing of the capacitor in RC circuit,
Charge at time t is q = Q[1- e -t/RC]
Where Q = maximum charge
qC = QC [1-e -t/RC ]
We know qC = V and QC = Vo Where Vo = maximum voltage
Therefore V = Vo [1- e -t/RC]
= 3 [1- e -(1.1 x10^-3 )/(300)(6x10^-6) ]
= 3[1- e -0.833]
=3(1- 0.6111)
=3(0.3888)
= 1.1666 volt

