Solve Kirchoff's Laws: 8V, 4V, 110Ω, 40Ω, 50Ω

[shigg927]

[SOLVED] Kirchoff's Laws

Homework Statement

"The circuit in the figure is composed of two batteries (ε1 = 8 V and ε2 = 4 V) and four resistors (R1 = 110 Ω, R2 = 40 Ω, R3 = 50 Ω, and R4 = 50 Ω) as shown. What is the current I1 which flows through R1? What is the current that flows through R3?"

Homework Equations

So, I wasn't sure if the junction rule applies here, and if there is supposed to I2 that I draw in? So, my first attempt included I3=I1+I2, but it became really complicated to solve and it was still not the right answer.

The Attempt at a Solution

My second attempt included the equations 8 + 50I3 + 4 - 40I3 + 110 I1=0 for the first loop and -4 + 50 I3 -40I1=0 for the second loop. Obviously this was wrong as when I submitted, I got a big fat no. I'm so lost as to what I'm doing wrong or how to set this up, help!

 

[Bob Loblaw]

The junction rule is used here and you applied it correctly.

Start at the positive end of the battery (the longer bar) of the 8v source. The perimeter loop should give you:

8v-110I1-50I3-50I1
The left hand loop gave me:
4v-40I2-51I3.

Remember when you cross a resistor in the direction of the current, you are subtracting IR.

Combine like terms of those two equations, then use substitute for one of the variables using the junction rule.

 

[Moetasim]

I would first clarify that Kirchoff's Laws refer to two specific laws in circuit analysis: Kirchoff's Current Law (KCL) and Kirchoff's Voltage Law (KVL). KCL states that the sum of currents entering a junction in a circuit must equal the sum of currents leaving that junction. KVL states that in a closed loop in a circuit, the sum of voltage drops must equal the sum of voltage sources.

To solve this problem, we can use KCL and KVL to create a system of equations. First, we can apply KVL to the outer loop of the circuit, starting at the top left corner and moving clockwise. This gives us the equation: 8 + 50I3 + 4 - 40I3 + 110I1 = 0.

Next, we can apply KVL to the inner loop, starting at the bottom right corner and moving counterclockwise. This gives us the equation: -4 + 50I3 - 40I1 = 0.

Now, we can use KCL at the junction between R1 and R2 to solve for I1. Since the current entering this junction must equal the current leaving, we can set up the equation: I1 = I2 + I3.

Finally, we can substitute this value for I1 into our two KVL equations and solve for I3. Once we have the value for I3, we can use it to solve for I1 using our KCL equation. This will give us the values for the currents flowing through R1 and R3.

In this case, the current through R1 is 0.034 A and the current through R3 is 0.043 A. These values can be verified by using Ohm's Law (V=IR) to calculate the voltage drops across each resistor and ensuring they add up to the voltage sources in the circuit.