Consider the circuit shown below. Find V1, I2, and I3. I1 =...

Transcript

00:01 In this problem, we have been given the following circuit and we need to determine from this circuit the value of v1, i2 and i3.

00:10 So we have weven as the potential difference applied by the cell here.

00:15 I2 and i3, these are the currents flowing through 6 -oom and 5 -oom resistor respectively.

00:22 So as we can observe that at this point, we apply kirchof's current law or junction rule, so that total current which is coming to this junction that is i2 plus 2 and this should be equal to the total current which is leaving off the junction that is i3 so we get this as the equation 1 and now let's apply kerchief's voltage law or loop law in this part 1 of the circuit so applying kirchap's voltage law we observe when we take the potential here as 0 we're going to get the potential here as we and there will be dropping the potential across this 12 -oom resistor and the potential at this point will be v1 minus 12 times 2 according to oms lock we is equal to ir that's v1 minus 24 and that will be the potential even at this point because there is no component in between but as we are moving in the clockwise direction across the 6 -oam resistor there will be potential drop by minus 6 into i2 so basically the current is in a opposite direction so we can say there will be potential gain here so we 1 minus 24 plus 6 i 2 that's equal to 0 this is the equation that we get let's mark it as equation 2 and now we apply kirchhoff's voltage a lot to this part of the circuit and we move in the clockwise direction even here in this part 2 of the circuit so applying in the similar manner we're going to get 21 minus 5 i 3 minus 6 i2 that is 0 so let's mark this as equation so we observed that we have the two equations and three variables so we can take one more loop so let's take the outer loop which includes v1 12 -oom resistor 21 volt battery and 5 -oom resistance so in this case when we apply kirchhoff's voltage law by moving in the clockwise direction we're going to get here the equation coming out to be v1 minus 12 into 2 that's 24 plus 21 minus 5 i 3 and this is 0 so let's mark this as equation 4 so we have four equations let's try to solve these four equations so first we can add these two equations so when we add these two equations we're going to get so we're going to anyways going to get this equation four only so let's take equation two three and one and let's solve so from equation one we get i3 as i plus i2 plus two let's substitute that in equation three so we're going to get 21 minus 5 times i 2 plus 2 minus 6 i2 is equal to 0 so from here let's solve and get the value i 2 so we're going to get 21 minus 5 i 2 minus 10 minus 6 i2 and that's 0.

03:34 So from here we're going to get 11 minus 11 i2 equals 0...

Question

Consider the circuit shown below. Find V1, I2, and I3. I1 = 2 A I2 I3 R1 = 12 ? R2 = 6 ? V2 = 21 V V1 R3 = 5 ? Hint V1 = V I2 = A I3 = A

Question

Consider the circuit shown below. Find V1, I2, and I3. I1 = 2 A I2 I3 R1 = 12 ? R2 = 6 ? V2 = 21 V V1 R3 = 5 ? Hint V1 = V I2 = A I3 = A

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