# Kirchhsoff's Circuit Law (Loop Rule Equation)

1. Apr 20, 2013

### Pruddy

1) Please take a look at the attachment below in order to solve the problem. Using the figure at the right determine the values of i1, i2 and i3. Input your answers in the form "a.bc x 10^(y) A".

2) Using the correct results from #1, determine the following. For A and B you must also tell if the current directions shown are correct. Give your answer in the requested format.

A. The current in the 10.0 resistor. Answer format: "abc mA, Y/N" . Y/N means choose Y if the current is as shown, and choose N if the direction of the current is not correct.

B. The current in the 4.5 V battery. Answer format: "abc mA, Y/N" . Y/N means choose Y if the current is as shown, and choose N if the direction of the current is not correct.

C. The voltage across the 15.0 resistor. Answer format: "ab.c V".

D. How much power is dissipated in the 5.0 resistor? Answer Format: "abc mW"

3. The attempt at a solution

Loop: i1(10Ω) - 9.0V + 6.0 V - i3(15Ω+20Ω) = 0
Loop: i3(15Ω+20Ω)-i2(5Ω+12Ω) - 4.5V = 0
Loop: i1(10Ω) - 9.0V + 6.0V + i2(12Ω + 5Ω) - 4.5V = 0
Junction: i1 + i3 = i2

Please I would like to verify if my loop equations are correct and also to know how i can solve for i1, i2, and i3. Thanks a lot!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### phys2.1.PNG
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2. Apr 20, 2013

### ehild

You have a sign error in the second loop equation. It has to be +.
Not all loop equations are needed. Only two of them are independent. Using two loop equations and the equation for the junction, you can solve for all currents.

ehild

3. Apr 20, 2013

### SammyS

Staff Emeritus

Loop: i3(15Ω+20Ω)-i2(5Ω+12Ω) - 4.5V = 0​
is incorrect.

It would be correct if i3 was in the opposite direction.

To solve:
You only need two (any two) of the loop equations. (The third is redundant.)​
Use the junction equation to plug in (i1 + i3) for i2, everywhere.

You then have two equations in two unknowns. Use Algebra skills to solve that.

4. Apr 20, 2013

### Pruddy

I used the two loop equations to solve for i3, i2, and i1. Please Take a look at my work below....

i1(10Ω) - 9.0V + 6.0 V - i3(15Ω+20Ω) = 0
i1 = (i3(15Ω+20Ω) + 3)/10

i1 = 0.3 + i3(3.5) ...................eqn 1

i3(15Ω+20Ω)+i2(5Ω+12Ω) - 4.5V = 0
i2 = (4.5 - i3(15Ω+20Ω))/(5Ω+12Ω)

i2 = 0.26 - i3(2.06) ................eqn 2

i1 + i3 = i2

0.3 + i3(3.5) + i3 = 0.26 - i3(2.06)

i3(3.5) + i3 + i3(2.06) = 0.26 -0.3

(6.56) i3 = -0.04

i3 = -0.04/(6.56) = -6.10 x 10^(-3)

i1 = 0.3 + (-6.10 x 10^(-3))(3.5) = 2.79 x 10^(-1)

i2 = 0.26 - (-6.10 x 10^(-3))(2.06) = 2.73 x 10 ^(-1)

These are the answers I got for i1, i2, and i3. Please, will you check to see if my calculations and answers are correct. Thanks a lot!!!

5. Apr 20, 2013

### SammyS

Staff Emeritus
I suggest keeping your numbers in exact form, by using fractions, or at least increase the accuracy by including more digits.

Notice here that you are taking the difference of two numbers which are nearly equal.

That can produce a percent error that's rather sizable, because you have rounded off.

You can check the answers by substituting into the loop equations.

I get an answer for i3 of i3 = -6/170 A ≈ -0.005381 A

6. Apr 20, 2013

### Pruddy

using substitution to solve i2 and i1, I get

i3 = -6/170 A ≈ -0.005381 A

i1 = ( -0.005381(15Ω+20Ω) + 3)/10 = 0.2811665

i2 = (4.5 - (-0.005381 (15Ω+20Ω))/(5Ω+12Ω)) = 0.2757844

using the information above, Please how do I calculate the current in the 10.0 resistor and How do i know the direction of the current. Thanks a lot... I appreciate.

7. Apr 21, 2013

### Pruddy

using substitution to solve i2 and i1, I get

i3 = -6/170 A ≈ -0.005381 A

i1 = ( -0.005381(15Ω+20Ω) + 3)/10 = 0.2811665

i2 = (4.5 - (-0.005381 (15Ω+20Ω))/(5Ω+12Ω)) = 0.2757844

using the information above, Please how do I calculate the current in the 10.0 resistor and How do i know the direction of the current. Thanks a lot... I appreciate.