8 2 2 points details my notes previous answers ask your...

Name: 8 2 2 points details my notes previous answers ask your teacher practice another consider the circuit shown in the figure below r 110 omega 100 omega 250 v 100 omega a b 500 omega 500 omega 74747
Uploaded: 2022-04-12T14:42:07-08:00
Duration: 5 min 53 s
Channel: Darshan Maheshwari
Description: 8 2 2 points details my notes previous answers ask your teacher practice another consider the circuit shown in the figure below r 110 omega 100 omega 250 v 100 omega a b 500 omega 500 omega 74747

8. [2 / 2 Points] DETAILS MY NOTES PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER Consider the circuit shown in the figure below. ($$R = 11.0 \Omega$$) 10.0 $$\Omega$$ 25.0 V 10.0 $$\Omega$$ a b 5.00 $$\Omega$$ 5.00 $$\Omega$$ R (a) Find the current in the $$R = 11.0 \Omega$$ resistor. 0.338 A (b) Find the potential difference between points a and b. 5.40 V

Transcript

00:01 This question, we have to consider the circuit and we need to find a current in this 18 -oom r is 18, 18 -oam resistor and the potential difference between a and v.

00:08 All right.

00:10 Okay.

00:11 So i think first let's try to club these two resistors because they are in parallel, right? so if we club them up, then its effective resistance is going to be 10 times 5 over 10 plus 5.

00:22 10 times 5 is 50.

00:24 So 50 over 15 is nothing but we just grab my calculator.

00:29 Is 33 .33.

00:31 It's 3 .3 .3.

00:32 Not 33 .3 .3 .3.

00:33 3 .33.

00:35 So the effective circuit is going to look like this.

00:39 This is, in fact, and these two resistance are also in series, right? are in series.

00:45 So 15 plus 18 is what? because are effective in case of series is just algebra.

00:49 5 plus 18 is 23.

00:51 So we can reduce these two resistances by a single resistance of 23 oms.

00:58 Over here we have this as one of the resistance and in parallel we have another 10 -oam along with the battery over here and this is how the circuit is being reduced now so this is 3 .33 ooms this is 10 and this is 25 volts this is 25 volts so if this is the case then let's let's try to analyze the circuit let's say this is i1 and this is i2 so the current will flow through here right will flow through here let's call it a high 3 and let's call this point as a so at a i'm going to apply the kcl and kvl if i apply kvl then one of the loop is like this one of the loop is like this and at a i'm going to apply the kcl so the incoming current is equal to the outgoing current so the equation is going to look like i -1 which is the incoming current should be equal to i -2 plus i -3 that's equation number one let's call it loop 1 and let's call it loop 2 so in the loop 1 the kvl is going to look like negative 23 i2 because it's going along the current then opposite to the current is 3 .33i3 is equal to 0 right so if we rearrange if we rearrange in fact if you notice carefully rather than doing all this don't you think that 33 .33 and this 23 are again in parallel to each other.

02:35 We could have done that and found the effective current.

02:38 But okay, fine, let's just do like this so that we can get the value of i2...

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