(a) Find the equivalent (total) resistance of the circuit....

(a) Find the equivalent (total) resistance of the circuit. [0.64 $\Omega$] Find the potential difference across R1, R2 and R3. [6.0 V; 6.0 V; 6.0 V] Find the current through each resistor. [3.0 A; 2.4 A; 4.0 A] Is total current of the circuit is same as the summation of individual current through each resistor? 3C-(b)Find the equivalent resistance of the following circuit, and the current supplied by the battery. [400 $\Omega$; 0.03 A] 12 V 600 $\Omega$ 400 $\Omega$ 800 $\Omega$ 560 $\Omega$

Transcript

00:01 So here's this circuit diagram.

00:02 It gives us values for r1, r2, and the battery voltage.

00:07 And we're essentially working our way down to find the equivalent resistance of the circuit and then find the total current through the battery, the potential difference across r2, the potential difference across r3.

00:20 I'm sorry, the 3 -on resistor, and then the current that runs through this 3 -home resistor.

00:27 So part a says find the equivalent resistance of r1 and 5 oom resistor.

00:34 So we're looking at this set here.

00:37 These two resistors are in parallel.

00:41 So for part a, we to find the equivalent resistance, we just add those two resistors inversely to each other.

00:47 So we have one over r1, which is 12, plus one over five.

00:55 And then we take the inverse of that.

00:58 So enter this into your calculator, and we should get a of 3 .53 oms.

01:06 So that's the answer to part a.

01:09 Now part b says knowing that we replace these two resistors with this resistor of 3 .53 oms, what is the equivalent resistance of this equivalent resistance plus this 4 ome resistor? so this equivalent resistance is going to be in series with this 4 oom resistor.

01:30 So we just add those straight through.

01:33 It's going to be the equivalent resistance of r1 and this 5 -oom resistor, which is 3 .53, plus this 4 -oom resistor.

01:43 So for part b, the equivalent resistance of this top branch of this circuit is 7 .53 oms.

01:54 Part c says now we've replaced these three resistors with this 7 .53 -oom resistor, and we want to find the equivalent resistance of this top branch to this 3 -on.

02:06 Resistor.

02:08 So part c, we have these two resistors in parallel.

02:13 So this 7 .53, the top branch of the circuit, is parallel to this 3 -on resistor.

02:19 So we add them just like we did in part a.

02:22 So 1 divided by 7 .53 plus 1 divided by this 3 -on resistor.

02:29 And then we take the inverse of that.

02:32 So the equivalent resistance of this section of the circuit, not including r2, is going to be 2 .15 oms.

02:52 So now we have essentially a circuit with just two resistors.

02:56 We have r2, and then we're going to take out this branch, this entire loop, and replace it with a resistor of 2 .15 oms.

03:04 So for part d, we can find the equivalent resistance for the entire circuit by just adding r2 to the equivalent resistance of this, which is just 2 .15.

03:16 So the problem told us that r2 was 1 .45 oms, and then we add that because it's in series to this section of 2 .15.

03:30 So we get an equivalent resistance for the entire circuit to be 3 .6 oms.

03:38 Now to find the current through the battery, we use oms law.

03:43 Now we know the total resistance for the circuit.

03:47 We know the total voltage that the circuit has.

03:52 And we can find the current by just manipulating oms law.

03:56 So oms law is v equals i r.

03:58 The voltage, the battery voltage equals the battery current times the total resistance of the circuit.

04:05 We're just going to manipulate this to solve for the total current.

04:08 So we'll divide over by r...

Question

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