Consider the circuit in the diagram. (a) Draw the simplest...

Consider the circuit in the diagram. (a) Draw the simplest equivalent circuit and label the values of the resistor(s). (b) What current flows from the battery? (c) What is the potential difference between points $A$ and $B ?$ (d) What current flows through each branch between points $A$ and $B$ ? (e) Determine the power dissipated in the $50.0-\Omega$ resistor, the $70.0-\Omega$ resistor, and the $40.0$ $\Omega$ resistor.

Transcript

00:01 In this problem, we have been given the following circuit and the resistors are as indicated here, which are connected.

00:09 And we have to first draw a simplified form of the circuit.

00:14 So here, in order to simplify the circuit, let's combine all the resistors.

00:18 So first we observe that these two resistors and these two resistors in the opposite branch, they are connected in series.

00:25 So first we just add them because that will be the total resistance of their combination.

00:31 So when we add the resistors in the upper branch, we're going to get 60 om here.

00:36 And adding the resistors in the lower arm, it will be 90 oms.

00:41 And here we observe 60 om and 90 om, they are in parallel.

00:45 So we applied the parallel combination of resistance to get the equivalent resistance out of these two branches.

00:52 So for that, we multiply 60 with 90, and we divide it with 60 plus 90.

00:58 And that's going to get us 36 om.

01:01 And now we observed that between a and b, 36 -oom resistor is the equivalent resistance of all these four resistors connected.

01:11 And that resistor is connected in series with this 50 -oom and 20 -oom resistance.

01:18 So we add all of them because they are in series.

01:21 So the total resistance will be 36 -oom plus 50 -oom plus 20 -oom.

01:28 And that will be 106 -oom.

01:30 So we have the total equivalent resistance of the circuit that's 106 om which is connected across 120 volt supply.

01:41 So this is the simplified form of the given circuit.

01:45 And now we have to determine the current which is coming out of the battery.

01:49 So for that, let's make use of omslaw.

01:52 We is equal to ir.

01:53 So the potential difference supplied by the battery is 120.

01:56 I is the current here and r is the equivalent resistance.

02:00 So in order to figure out the current, let's divide 120 with 106.

02:05 And that's going to get us the value of current which is coming out of the battery as 1 .13 amps approximately.

02:14 And in the next situation, we have to determine the difference in potential between the points a and b.

02:22 So in order to compute the difference, we observed that the total potential difference supplied to the circuit is 120 volt.

02:29 And from this or out of this potential difference which is supplied some potential difference is dropped by this 50 -oom and 20 -oom resistor before coming to this point a...

Question

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