ElEn 236
1.
voltage
source
Ohm’s Law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law
Symbols
resistor
ground
cap
inductor
signal
source
volt
meter
current
meter
V
2.
I
Conventional current ( dc )
Even though current is really electrons moving in the direction of out from
the negative terminal of the source, through the circuit and back through the
positive terminal, we will follow the convention that current is flowing
from the positive terminal of the source, through the circuit and back through
the negative terminal of the source.
3.
Ohms Law
The amount of current in amperes that flows through a resistor is equal to the V
voltage in volts that is applied across the resistor divided by the resistance of
the resistance in Ohms.
I = V / R so V = IR and R = V / I
eg if the voltage drop across a 200 Ω resistor is 12 V then the current through
it will be 0.06 A or 60 mA
4.
Resistors in Series ;
I
R
Kirchoffs voltage law
R1
The applied voltage must equal the sum of the voltage drops. That is, the sum
of the voltage drops across R1 and R2 must equal the applied voltage E .
Therefore the total resistance seen by the applied voltage is R1 + R2 .
So I = E / ( R1 + R2 )
eg if : R1 = 2 kΩ , R2 = 3 kΩ and E = 20 V then I = 4 mA
5.
I
E
R2
Voltage Dividers
In the case above, the voltage drop across R1 would be 8 V ( V = I R ) and the voltage drop
across R2 would be 12 V . VR1 = E ( R1 / ( R1 + R2 ) )
I
6.
Voltage and Current Measurement
R1
An ammeter must be placed in series with the circuit while a voltmeter must
be placed across the device whose voltage drop is to be measured.. In order
to measure the voltage and current for R1 in the previous case, the meters
would have to be placed as shown.
7.
E
R2
I
Resistors in Parallel ;
Kirchoffs Current Law
The total current will equal the sum of the parallel currents
E
R1
I1 R2
I2
V
The parallel components will have the same voltage drop
So I = I1 + I2
The resistance ( RT ) seen by the source E will be :
RT = 1 / ( 1/R1 + 1/R2 )
eg
if R1 = 1 kΩ , R2 = 2 kΩ , E = 10 V then :
I1 = 10 mA , I2 = 5 mA , I = 15 ma
RT = 1 / ( 1 / 1 kΩ + 1 / 2 kΩ ) = 666.7 Ω
I = 10 V / 666.7 Ω = 15 mA
8.
Current Dividers
If the total current is known , and there are only two parallel resistors , then a short-cut way tofind
The current through one of the resistors is : I1 = I ( R2 / ( R1 + R2 ) )
eg
in the circuit above I1 = 15 mA ( 2 kΩ / 3 kΩ ) = 10 mA
9.
Series and Parallel Symbols
If R1 and R2 are in series , we write : R1 + R2
If R1 and R2 are in parallel , we write : R1 // R2
10.
Power
Power in Watts is equal to the voltage across a device in Volts multiplied by the current through the
2
2
device in Amps. i.e. P = VI . If the device is a resistor then P = V / R or P = I R
For AC voltages or currents, if they are in rms equivalents , then we can used the DC equations .
11.
Resistors in Series / Parallel Circuits
In the circuit shown , the resistance ( RT ) seen by the source ( E )
is R1 + ( R2 // R3 ) or :
RT = R1 + 1/( 1/R1 + 1/R2 )
Eg if R1 = 1 kΩ , R2 = 3 kΩ , R3 = 6 kΩ , E = 12 V , then :
I
R1
E
R2
R3
R2 // R3 = 1 / ( 1/3000 Ω + 1/6000 Ω ) = 1 / ( 333.3 μS + 166.7 μS )
= 1 / 500 μS = 2000 Ω
RT = R1 + ( R2 // R3 ) = 1 kΩ + 2 kΩ = 3 kΩ
IT = 12 V / 3 kΩ = 4 mA
VR1 = 4 mA x 1 kΩ = 4 V , V R2 // R3 = 4 mA x 2 kΩ = 8 V , or V R2 // R3 = 12 V – 4 V = 8 V
IR2 = 8 V / 3 kΩ = 2.667 mA , IR3 = 8 V / 6 kΩ = 1.333 mA
IE = 4 mA , IR1 = 4 mA , IR2 + IR3 = 4 mA
PE = 4 mA x 12 V = 48 mW
PR1 = 4 mA x 4V = 16 mW, PR2 = 2.667 mA x 8V = 21.333 mW, PR3 = 1.333 mA x 8V = 10.667 mW
PR1 + PR2 + PR3 = 48 mW
PE = PR1 + PR2 + PR3