Refer to the circuit diagram on the board. Assume that the switch has been close
ID: 1836289 • Letter: R
Question
Refer to the circuit diagram on the board. Assume that the switch has been closed for a long time (ever since t = - infinity.) How much current is flowing through the battery? What is the charge on the capacitor? Eventually t=0 arrives and the switch is opened. How much current initially flows through the 6-Ohm resistor just after the switch is opened? Write down a differential equation that involves the current from the capacitor, its charge, and any resistance involved for t=0 and after. Solve the equation for q(t) and graph the voltage on the capacitor as a function of time. When, if ever, will the voltage on the capacitor be equal to 1 00 V?Explanation / Answer
2 a) the capacitor is open circuit
therefore the circuit acts as a simple circuit with a battery of 24 V and two branches in parallel
one branch contains 4 ohm and 8 ohm in series , net resistance = 8 + 4 = 12 ohm
other branch contains 2 ohm and 6 ohm in series , net resistance = 6 + 2 = 8 ohm
net resistance of the circuit = 8 || 12 = 4.8 ohm
current through the circuit at t= infinity is 24/4.8 = 5A
b)
the 5A current divides into two branches
current through the 12 ohm branch = I1
current through the 8 ohm branch = I2
I1 * 12 = I2 * 8 = 24
therefore I1 = 2A , I2 = 3A
there potential of the end of capacitor near 4 ohm resistor = 24 - I1 *4 = 16 V
there potential of the end of capacitor near 6 ohm resistor = 24 - I2 * 6 = 6 V
therefore total potential difference across the capacitor = 16 - 6 =10V
charge on the capacitor = C *V = 2*10-6 * 10 = 2*10-5
c)
as switch is opened at t=0
capacitor acts as a battery with voltage 10 V with two branches
one branch contains 4 ohm and 6 ohm in series , net resistance = 10 ohm
other branch contains 8 ohm and 2 ohm in series , net resistance = 10 ohm
therefore let current through 6 ohm at t=0 be I3
I3 * 10 = 10 V => I3 = 1 A
d)
IF we consider kirchoffs voltage law in the loop containing capacitor 6 ohm and 4 ohm we get
Vc - I/2 * 6 - I/2 *4 = 0
Vc = 5I [ Vc = voltage across capacitor = Q/C I = dQ/dt ]
Q/C =5 dQ/dt
=> Q/ (2*10-6) = 5 dQ/dt this is the differential equation