Kirchhsoff's Circuit Law (Loop Rule Equation)

[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"

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!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[ehild]

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".
...

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!

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

 

[SammyS]

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"

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!

Your 2nd Loop equation

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.

 

[Pruddy]

Thanks a lot for your reply.
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!

 

[SammyS]

Thanks a lot for your reply.
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

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

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

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.

(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!

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

 

[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.

 

[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.