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[SOLVED] RC Circuit Help
1. Homework Statement
What is the potential of point a with respect to point b in the circuit shown when switch S is open? Which point, a or b is at a higher potential? b.) What is the potential (after a long time) of point a with respect to ground when switch S is closed? How much does the charge in each capacitor change when S is closed?
R1 = 6 Ohms, C1 = 6microFarads, C2 = 3 microFarads, R2 = 3 ohms, C3 = 6 microFarads
V = 18V
Im sorry for the poor drawing
2. Homework Equations
V = IR
C = Q/V
3. The Attempt at a Solution
Ok, I know the solution to this question, I just don't know why...
A.) Apparently this circuit simplifies to a circuit of just the capacitors, the resistors drop out and you end up with C1 and C2 in parallel, connected to C3. Why do the resistors drop out?
From this you can combine all 3 resistors into 1 resistor Ctotal = 3.6 microFarad, and so
Q = CV = (3.6microF)(18V) = 6.48x10^5
then combining C1 and C2 to get C12 = 9microF, i can get the voltage drop accross the capacitors = Q/9microF = 7.2V
B.) When the switch is closed, aparently all the voltage drop is accross the bottom capacitor, C3... why???
Thank you very much for any help
1. Homework Statement
What is the potential of point a with respect to point b in the circuit shown when switch S is open? Which point, a or b is at a higher potential? b.) What is the potential (after a long time) of point a with respect to ground when switch S is closed? How much does the charge in each capacitor change when S is closed?
R1 = 6 Ohms, C1 = 6microFarads, C2 = 3 microFarads, R2 = 3 ohms, C3 = 6 microFarads
V = 18V
Im sorry for the poor drawing
2. Homework Equations
V = IR
C = Q/V
3. The Attempt at a Solution
Ok, I know the solution to this question, I just don't know why...
A.) Apparently this circuit simplifies to a circuit of just the capacitors, the resistors drop out and you end up with C1 and C2 in parallel, connected to C3. Why do the resistors drop out?
From this you can combine all 3 resistors into 1 resistor Ctotal = 3.6 microFarad, and so
Q = CV = (3.6microF)(18V) = 6.48x10^5
then combining C1 and C2 to get C12 = 9microF, i can get the voltage drop accross the capacitors = Q/9microF = 7.2V
B.) When the switch is closed, aparently all the voltage drop is accross the bottom capacitor, C3... why???
Thank you very much for any help
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