Chapter 18Kirchhoff’s Rules

RC Circuits

1) The sum of the currents entering a junction must equal the sum of currents leaving a junction. (Conservation of Charge)

2) The sum of potential differences across all the elements on any closed loop in a circuit must be zero. (Conservation of Energy)

Junction Rule

Loop Rule

2

2

10

5

3I

I3

I2

I1 I

VThe Junction Rule says that

I = I1 + I2 + I3

2

2

10

5

3I

I3

I2

I1 I

VWe can use the loop rule multiple times:

V - 3 I - 10 I2 - 2I = 0 V - 3I - 2I3 - 2I = 0

V - 3 I - 5 I1 - 2I = 0

We now have a system of 4 equations and 4unknowns. We can therefore solve for eachof the 4 unknown currents.

V - 3 I - 10 I2 - 2I = 0 V - 3I - 2I3 - 2I = 0

V - 3 I - 5 I1 - 2I = 0I = I1 + I2 + I3

IV I

1 5

5

IV I

3 5

2I

V I2

5

10

Plug these three intothe current equation.

IV I V I V I

5

5

5

10

5

2

10I = 8V - 40 I

I = 0.16 V

Plug this back into the other 3 equations...

I1 = 0.04 V I2 = 0.02 V

I3 = 0.1 V

For Applying Kirchhoff’s Rules

A) assign a direction to the current in eachbranch of the circuit. Just GUESS!! If yourguess is incorrect, the current will come outas a negative number, but the magnitude willstill be correct!

B) when applying the loop rule, you mustchoose a consistent direction in which to proceed around the loop (either clockwise or counterclockwise, your choice, but stick to it).

1) When you encounter a resistor in the directionof the current, the voltage drop is V = - I R

2) When you encounter a resistor opposite thecurrent, the voltage drop is V = + I R

R2

I

R1 I1

I2 - I1R1

+I2R2

3) When you encounter an emf in the directionyou’re going around the loop, the voltage change is +V

4) When you encounter an emf opposite thethe direction you’re going, the voltage change is -V

- V1

+ V2

V1 V2

5) The junction rule can only be applied n-1times in a circuit with n junctions.

6) Each new equation you write must containa current that you haven’t yet used.

7) To solve a system of equations with kunknown quantities, you need k independentequations.

2

2

10

5

3I

I3

I2

I1 I

V

This circuit has two junction points. Therefore,we can only use the junction rule ONE time.

Find the currents in each branch of this circuit.

I0 I1

I26

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

There are two junction points, so we can applythe junction rule ONCE.

I0 = I1 + I2

6

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

How many possible loops are there?

b c d a e f c d a b e f c b

We have 3 unknowns, so we need 3 equationstotal. Therefore, we need use only 2 of the 3equations provided by the loop rule. (The junctionrule gave us 1 equation already!)

I0 6

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

I1

I2

Let’s go around bcda first.

-(6) I1 + 18 V - (12) I0 = 0

(12) I0 + (6) I1 = 18 V

I0 6

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

I1

I2

Let’s go around efcb second.

12 V - (1) I2 + (6) I1 - (1) I2 = 0

(2) I2 - (6) I1 = 12 V