PHY 202 Greenville Tech Kirchhoff's Rules Name Show your...

PHY 202 Greenville Tech Kirchhoff's Rules Name Show your work. 1.) Find the current in each branch of the network using Kirchoff's rules. R1 = 5 ?, R2 = 2 ?, R3 = 3 ?, ?1 = 10 V, and ?3 = 30 V R1 R2 R3 ?1 ?3 E

Transcript

00:01 So here we are given a circuit in which the resistant r1 is 5 oom, resistance r2 is 2 oom, resistance r3 is 3 oom, battery supply e1 is 10 volt, battery supply e3 is 30 volt.

00:18 So we have to find the current flowing through the circuit by using the k -chops route.

00:23 So let us talk about the concept which we are going to use here is the k -h -oves, voltage law.

00:31 According to khrop's voltage law, the sum of the voltage inside of the loop is always request to zero.

00:38 So let us look to the figure first.

00:40 So this is what the figure which we are given here according to question.

00:44 So let's suppose the current flowing throughout this branch is i1 and current flowing throughout this loop is i2.

00:52 Let's suppose this has first loop, this as second loop.

00:56 So we will say applying kvl to loo first.

01:02 So as we are going in the same direction as the current flowing, so in that case, now we are going towards the positive of the battery supply, that's why it will be positive e1, minus o.

01:13 We are going in the same direction as the current flowing, that's why it will be i1 times of r1, minus of.

01:21 Now here, the current flowing throughout the branch of r2 will be i1 plus i2.

01:26 So it will be simply i1 plus i2 times of r2, this should be equals to 0.

01:32 So on substituting the values, e1 is 10 volt minus i1, we have to find, r1 is 5 minus of r2 is 2.

01:44 So we can directly open the bracket, so 2 times of i1 minus 2 times of i2 will be equals to 0.

01:51 So if we form the equation here.

01:53 So what we get 5 plus 2 will give me.

01:56 7 times of i1 plus 2 times of i 2 this should be quers to 10 let this as equation number 1 now similarly we will apply the kvl to 2 second loop so in that case what we get that is e3 minus of i2 times of r3 minus of i 1 plus i 2 times of r 2 this should be quast to z now on substituting the values, i3 is 30 volt minus of i2 we have to find here r3 is 3 ohem minus of i1 plus i2 times of r2 is 2 is equal to 0.

02:42 Now if we form the equation here.

02:44 So 3 i2 minus 2 i2 will become 5 times of i2 here while we are having 3 times of i1.

02:53 So here it will be 3 times of i1...

Question

PHY 202 Greenville Tech Kirchhoff's Rules Name Show your work. 1.) Find the current in each branch of the network using Kirchoff's rules. R1 = 5 ?, R2 = 2 ?, R3 = 3 ?, ?1 = 10 V, and ?3 = 30 V R1 R2 R3 ?1 ?3 E

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