P4 referring to the given figure assume r0 2 omega r1 3...

P4. Referring to the given figure, assume $R_0 = 2 \Omega$, $R_1 = 3 \Omega$, $R_2 = 4 \Omega$, $R_3 = 5 \Omega$, and $v_S = 10 V$. Use KCL and Ohm's law to find the unknown currents. (Round the final answers to two decimal places.) A B $R_0$ $i_0$ $i_1$ $v_S$ $R_1$ $R_2$ $i_2$ $R_3$ $i_3$ 6A

Transcript

00:01 Hi, so you're considering a circuit and you're asked to find the current through each resistor before and after closing the switch.

00:11 So before closing the switch, we can essentially ignore r2 because the switch is open.

00:18 There will be no current flowing through that branch of the circuit.

00:21 So i've not even drawn it on our diagram here.

00:26 And we see that sorry i just realized i labeled these incorrectly so i'm going to fix that really quickly first so as we're considering what the current is going to be i know that the current through the battery has to be the same as the current through r1 so i1 here is going to be the same as the current through the battery and i can figure that out by using omslaw b, e, e, equals ir with our equivalent resistance and the voltage of the battery.

01:05 So i know that i1 is going to be equal to v over r equivalent.

01:13 And now we need to figure out r equivalent.

01:16 R3 and r4 are in parallel with each other.

01:19 So i know that 1 over r3 ,4 is going to be equal to 1 over r3.

01:28 Plus 1 over r4, which will be 1 over 1113 plus 1 over 113.

01:36 And when you actually calculate that, you would find that r3 -4 is equal to 56 .5 oms.

01:52 But now if we were to redraw the diagram, it's now going to look like this, where this is our r34 that we just calculated and this is r1.

02:05 So those two are now in series.

02:07 So the total equivalent resistance is going to be r1 plus r34, which ends up being equal to about 169 .5 poms.

02:21 And when you plug that in, you get that the current i -1, then must be equal to 0 .133 amp.

02:31 So that tells us the current through resistor one, but i'm going to have to erase here to make some space.

02:41 In order to figure out the current through resistors 3 and 4, we want to sort of think about our kirkoff laws or our loop rules.

02:53 I know that that current i1 is going to be split into i3 and i4.

03:04 And so i1 is equal to i3 plus i4 but because the resistance and voltage drop across those resistors is going to be the same the currents have to be the same again by um um um olms law and so i1 is equal to two times i3 which means that i3 is half of i1...

Question

P4. Referring to the given figure, assume $R_0 = 2 \Omega$, $R_1 = 3 \Omega$, $R_2 = 4 \Omega$, $R_3 = 5 \Omega$, and $v_S = 10 V$. Use KCL and Ohm's law to find the unknown currents. (Round the final answers to two decimal places.) A B $R_0$ $i_0$ $i_1$ $v_S$ $R_1$ $R_2$ $i_2$ $R_3$ $i_3$ 6A

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