INC 112 Basic Circuit Analysis

Week 2

Kirchhoff's laws

Kirchoff’s Current Law (KCL)

“The sum of all currents entering a point is equal zero.”

I1

I4 I3

I2

I1 - I2 - I3 + I4 = 0 or

-I1 + I2 + I3 - I4 = 0

KCL Metaphor

From the pipe that is full of water, the amount of flow-in water must be equal to the amount of flow-out water.

This is because water cannot disappear.

How to use KCL

1Ω2Ω

5Ω

2A3V

Place a circle anywhere any size that you want

Kirchoff’s Voltage Law

“The sum of all voltages in a closed loop is equal zero.”

+

V1

-

+

V3

-

+ V2 -

V1 – V2 – V3 = 0 or -V1 + V2 + V3 = 0

KVL Metaphor

2 m

3 m

5 m

A

C

B 0)5()2()3(

0

ACCBBA hhh

How to use KVL

1Ω2Ω

5Ω

2A3V

Form a loop in the circuit

Example: Voltage Divider

10V

4K

6K

X mA

KCL: Current is the same at all points in all the circuit because there is no branch.

mAx

xx

xx

140006000

10

1040006000

04000600010

KVL:

Voltage Divider Circuit

10V

4K

6K

1mA

0V

10V 10V

4V

0V

Resistor Reduction

R1 R2

R2

R1

R1+R2

(R1*R2)/R1+R2

Series

Parallel

21

111

RRRtotalor

Example

10V 4K2K

7.5mA

0V

10V 10V

0V

10V

0V

2.5mA

5mA

Voltage & Current Sources Combination

3V

2V

1V

4V

6A 5A 4A 7A

Example: Current Divider

4K2K1mA

1mA

0V

1.333V 1.333V

0V

1.333V

0V

0.333mA

0.667mA

R1

R2

V

+VR1

-

+VR2

-

Voltage Divider

VRR

RV

VRR

RV

R

R

21

221

1

2

1

Current Divider

R2R1I

IR1 IR2

IRR

RI

IRR

RI

R

R

21

121

2

2

1

Example

10V

4K

2K2K

+V1-

Find V1

VVKK

KV 333.310

42

21

Circuitswith more than 1 ground

10V 2K10V 2K

Same circuit

Voltmeter and Ammeter

VVoltmeter: Measure the voltage between 2 pointsUsage: Put each pin on each point

AAmmeter: Measure the current that passes through a wireUsage: Cut that wire and connect the cut to the ammeter

10V

4K

6K

Voltmeter

V

Ammeter

10V

4K

6KA

Voltmeter and Ammeter

V

Same as electric wire

Property: R = 0 ohm

A

Same as open circuit

Property: Very high resistance

Dependent Source

The amount of voltage (current) supplied depends onother voltage (current).

• Dependent Voltage Source

• Dependent Current Source

+

-

4Ix

2Vx

Example

+-

5V

4Ω 1V 2Ω

3Vx

+ Vx -

I

Find I

AI

I

III

IVxVxII

222.0

418

0122145

)4(,032145