Consider the circuit shown below. (a) Find the current through each resistor. (b) Check the calc

step 1 In this question we have this particular interesting circuit. We have five different resistors a

Consider the circuit shown below. (a) Find the current...

Key Concepts

Power Calculation in Circuits

Power calculation in an electrical circuit involves determining the rate at which energy is used or dissipated, typically calculated using the formula P = VI, where V is the voltage and I is the current. This concept is important for verifying the correctness of circuit analysis, as the power supplied by sources should equal the sum of the power dissipated across circuit elements, ensuring energy conservation.

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law asserts that the total current entering a junction must equal the total current leaving the junction. This law is fundamental for circuit analysis, especially at nodes where multiple branch currents converge, as it provides the necessary equations to solve for unknown currents in various parts of the circuit.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits that relates voltage (V), current (I), and resistance (R) through the equation V = IR. It is essential for calculating the current through resistors when the voltage across them and their resistances are known, making it a key tool for analyzing linear circuit components.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law states that the sum of the electrical potential differences (voltage) around any closed network is zero. This concept is crucial when analyzing circuits with multiple loops, as it allows one to write equations that relate the voltage drops across different resistors and sources, ensuring the consistency of potential differences in the circuit.

Transcript

00:01 In this question we have this particular interesting circuit.

00:05 We have five different resistors and we want to find five different currents.

00:10 They are passing through each of the resistor.

00:15 So we're going to have to use kachov's loop rule as well as the junction rule in order to get five different equations that we will need to use to solve for the current.

00:30 Right now the current directions are arbitrary i've chosen them arbitrarily i don't know what is the actual direction but using this direction if we get a value or a negative value for the current it means that our choice of current direction is wrong and we will just you know take the opposite direction to be the actual current right so there's no issues with choosing and arbitrary direction first and then solving for the question.

01:07 So now let's do the junction rules which are the easiest.

01:13 We have two junctions, so junction one here and junction two.

01:20 For junction one, the incoming current is i1 and the outgoing current is i5 and i3.

01:32 For the second junction, the incoming current is i5 and i3.

01:35 Current is i5 and i2 and the outgoing is i 4 so this is the overall current now we actually have another current another junction rule but this is actually the across the entire circuit right so the from the entire circuits we know that this current over here i or the current passing through the battery, this equal on this side as well as this side.

02:21 So what we can tell is actually the third equation is i3 plus i4.

02:30 It's actually equals to i1 plus i2.

02:37 So now moving on to the loop rule.

02:43 Draw them in purple.

02:45 One of the loop could be passing through just the internal diamond shape over here.

03:07 We can have another loop that is passing at the top, sorry, in the opposite direction.

03:23 It would be better.

03:25 The top half of the diamond.

03:32 And then we can have an external loop.

03:34 Loop that is passing through i1 right so from i1 down to i3 and back so this is the three loops that we'll need now let us do for the equations for the loops right for the first loop which is the triangle below we start off with r4, since we are moving in the same direction as r4, with the current, we're moving in the same direction, it's current, so we have to subtract away the potential drop.

04:43 R4 .i4.

04:45 We are moving in the opposite direction to the current for r3, so we'll add r3, and lastly we are moving same direction as the current for r3.

05:04 So there will be minus r5 i5 goes to 0.

05:14 Fifth equation, the upper half of the diamond.

05:23 We start off with r1.

05:26 So r1, we are moving in the same direction as the current.

05:29 So we subtract r1, i1...