The capacitor in the figure is initially uncharged. There is...

The capacitor in the figure is initially uncharged. There is a switch not shown in the figure is in series with the emf and between it and resistor R1, the switch is then closed at t=0. a. Immediately after the switch is closed, what is the current through each resistor? b. What is the final charge on the capacitor? c. What is the current through each resistor after the switch has been closed for a very long time? $R_1 = 8.00 Omega$ $mathcal{E} = 42.0 V$ $R_3 = 3.00 Omega$ $R_2 = 6.00 Omega$ $C = 4.00 mu F$

Transcript

00:01 Hello, here we have to find current through each resistor right after the circuit is closed.

00:08 Now let's make the sketch here.

00:10 Let's start with the emf.

00:13 So current flows through r1, then it branches into two parts.

00:20 One goes through r2, another goes through r3, and it also charges the capacitor, and the resistance of the capacitor is very close to 0 oms, when it's been charged.

00:35 So, that's why we can basically omit it.

00:39 And here, we first let's find the equivalent resistance of this circuit, that is r1 plus r2, r3 over the sum of r2 and r3.

00:51 Now it's calculated, and that is 10 oms.

01:24 Now, once it's calculated, we can determine the current through r1.

01:30 Current through r1 is emf over the equivalent resistance, which is, 4 .20 ampires.

01:47 Now that is calculated and let's determine the voltage drop across r2 and r3 which are the same.

01:55 The emf minus i through r1 times r1.

02:12 That's 8 .40 volts.

02:16 And now we can calculate current through r2 that is 8 .40 volts over 6 homes.

02:32 That is 1 .40 ampires and now we can determine current through r3 which is 8 .40 volts divided by 3 oms that is 2 .80 ampers once it solves we can move on to question b where we have to calculate charge of the capacitor so here once it's charged voltage across the capacity can be found as following...

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