Circuits With Multiple Sources

• Suppose you have the following circuit diagram:• How do you determine the

current throughout the circuit?• How do you determine the

voltage throughout the circuit?• Cannot use equivalent

circuits to simplify this…what to do???

- -

Kirchoff’s Circuit Laws

• Gustav Kirchoff– generalized observations of circuits• Kirchoff’s Junction Rule (aka: Kirchoff’s

current law)• Current into node = Current leaving

node• Iin = Iout

• Kirchoff’s Loop Rule (aka: Kirchoff’s voltage law)• Sum of voltage around any closed

loop must be zero• Vloop = 0 V

Image obtained from: http://www-history.mcs.st-and.ac.uk/BigPictures/Kirchhoff_4.jpeg

Applying Kirchoff’s Laws

• Step 1: Decide on directions for current• Electron flow notation:

electrons move away from -, move toward +• Every node must have at

least 1 current going in and 1 going out• # currents = # wire lines

between nodes

- -I1

I1

I1

I2

I2

I3

I3

I3

Applying Kirchoff’s Laws

• Step 2: Sign conventions for components• Electron flow notation:

electrons move away from -, move toward +• Every component needs a

(-) end and a (+) end - -I1

I1

I1

I2

I2

I3

I3

I3

-

--+

+

+

Applying Kirchoff’s Laws

• Step 3: Perform Junction Rule for every node• Iin = Iout

• Node A: I1 + I2 = I3

• Node B: I3 = I1 + I2

• Good! We’re consistent! - -I1

I1

I1

I2

I2

I3

I3

I3

-

--+

+

+

A

B

Applying Kirchoff’s Laws

• Step 4: Perform Loop Rule for every loop• Vloop = 0 V• Going from - to +: negative

voltage (V being lost)• Going from + to -: positive

voltage (V being gained)• Remember: V = IR

- -I1

I1

I1

I2

I2

I3

I3

I3

-

--+

+

+

A

B

Loop 1 Loop 2

Loop 3

Applying Kirchoff’s Laws

• Loop 1:

-I1R2 - ε2 - I1R1 + ε1 = 0 V• Loop 2:

-I3R3 + ε2 = 0 V• Loop 3:

-I1R2 - I3R3 - I1R1 + ε1 = 0 V- -I1

I1

I1

I2

I2

I3

I3

I3

-

--+

+

+

A

B

Loop 3

Loop 1 Loop 2

Applying Kirchoff’s Laws

• Let’s say we have the following:• ε1 = 6.0 V• ε2 = 1.5 V• R1 = 2.0 Ω• R2 = 3.0 Ω• R3 = 5.0 Ω

What is the current passing through the battery ε2?

- -I1

I1

I1

I2

I2

I3

I3

I3

-

--+

+

+

A

B

Loop 3

Loop 1 Loop 2

Applying Kirchoff’s Laws

• Looking at diagram, I2 is current passing through ε2

• How to find I2?• I3 = I1 + I2 so I2 = I3 - I1

• How to find I3?• -I3R3 + ε2 = 0 V so I3 = ε2/R3

• How to find I1?• -I1R2 - ε2 - I1R1 + ε1 = 0 V

soI1 = (ε1 - ε2) / (R1 + R2)

What we know:I3 = I1 + I2

-I1R2 - ε2 - I1R1 + ε1 = 0 V

-I3R3 + ε2 = 0 V

-I1R2 - I3R3 - I1R1 + ε1 = 0 V

Applying Kirchoff’s Laws

• I3 = ε2/R3

• I1 = (ε1 - ε2) / (R1 + R2)

I1 = (6.0 V - 1.5 V) / (2.0 Ω + 3.0 Ω)

I1 = 4.5 V / 5.0 Ω

I1 = 0.90 A

• I2 = I3 - I1

What we know:I3 = I1 + I2

-I1R2 - ε2 - I1R1 + ε1 = 0 V

-I3R3 + ε2 = 0 V

-I1R2 - I3R3 - I1R1 + ε1 = 0 V

Oh no! A negative current! What does that mean?

= 1.5 V / 5.0 Ω = 0.30 A

= 0.30 A - 0.90 A = -0.60 A

Negative current: goes in opposite direction of what you said it did

Try Another!

Information:• ε1 = 120 V• ε2 = 120 V• R1 = 1.0 Ω• R2 = 2.0 Ω• R4 = 4.0 Ω• R5 = 5.0 Ω• R6 = 6.0 Ω

Find currents across: R1, R4

Let’s Step It Up A Bit!

Information:• ε1 = 120 V• ε2 = 120 V• R1 = 1.0 Ω R5 = 5.0 Ω• R2 = 2.0 Ω R6 = 6.0 Ω• R3 = 3.0 Ω R8 = 8.0 Ω• R4 = 4.0 Ω

Find currents across: R3, R6

19.0 A

15.9 A