Consider the following circuit diagram. 1) What is the current from A to B 2) What is the current from A to B if the switch was closed 3) What is the current from A to B when the switch closed after the capacitor is charged 4) What is the current from C to D if the switch is closed and the after the capacitor is charged 5) what is the current from b to e in the circuit after the switch closed and the capacitor charged

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anonymous · Answer

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anonymous · Answer

for 1) From A To B I assume it's just the equation for current with 12 volts and 4 ohms of resistivity so its 3 Amps?

anonymous · Answer

No. It's a bit more complicated. Let me walk you through it.

anonymous · Answer

Sorry. Internet issues.

anonymous · Answer

Here we go. 

Let's write as many Krichoff's voltage and current law expressions as we can to help us define the voltages and currents through each circuit element. 

$$KVL (a,b,e,f):~~ -12 + V_{4 \Omega} + V_{8 \Omega} = 0$$$$KVL(a,c,d,f):~~ -12 + V_{4 \Omega} + {1 \over C} \int\limits i_C dt = 0$$$$KCL(b):~~ - i_{4 \Omega} + i_{8 \Omega} + i_C = 0$$Remember that$$i_C =C {dV_C \over dt}$$and that once a capacitor is completely charged there is no voltage drop across it and all current will pass. 

If at t=0, the switch is open and the capacitor has no charge, there will be no voltage or current flow as we don't have a complete circuit. When the switch is closed, the capacitor will charge. Once fully charged (typically after 3 times constants) and the switch opened, the capacitor will discharge. 

Can you take it from here?

anonymous · Answer

So let me get this straight with the switch closed, The current from  A to B will be 0 amps, When the switch is open[and before the capacitor gets a charge], The current from A to B will be Voltage - Resistance[x] = 0 which in the case would be 1 amp?

anonymous · Answer

With the switch open, no current will flow. When the switch closes, the current from A to B will be $$i_{A/B} = {V_{battery} \over R_{4 \Omega}}$$

anonymous · Answer

Because we are just looking at a to b the 8 ohm resistor is ignored?

anonymous · Answer

That should be the case. You should use the Kirchoff's laws expressions to verify this.

anonymous · Answer

Whoops. The expression for current from A to B should be$$i_{A/C} = {V_{4 \Omega} \over R_{4 \Omega}}$$where $V_{4 \Omega}$ is the voltage drop across the 4 Ohm resistor.