Not sure how they got the answer.. For the circuit question In the Rc circuit sh

Question

Not sure how they got the answer.. For the circuit question In the Rc circuit shown to the right above the capacitor is initially uncharged. the switch is closed at t = 0. Immediately after the switch is closed, what is the current i_1 through the batter? You do not have to solve any differential equations to get the answer. [A e_1 = 0 [B] e_1 = 2 A [C] i_1 = 3 A [x] i_1 = 4 A [E] i_1 = 5 A In teh same RC circuit shown above, at the moment in time when the current i_1 = 3.5A, find the voltage V_c across the capacitor. You do not need to solve any differential equations to get the answer. [A] V_c = 12 V [B] V_c = 7 V [C] v_c = 4 V [X] V_c = 3 v [E] V_c = 2 V

Explanation / Answer

Initially the Capacitor is uncharged. As soon as the switch is closed, the voltage drop across the Capacitor rises from zero to peak value as the current flowing through it decreases with time. So just when the switch is closed, it is as if the capacitor is not there in the circuit.

So now we see that there are two 2ohm resistors in parallel. So its equivalent resistance will be:

1/Req = 1/2 + 1/2

therefore Req = 1 ohm

this is now in series with a 2 ohm resistor and so the final resistance in the circuit will be: 2 + 1 = 3 ohms

So when the voltage source is providing 12 volts, the current i1 when the switch is just closed will be: 12/3 = 4A.

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