Download Series and Parallel Circuits

Ohm’s Law
V = voltage, units are Volts (V)
I = current, units are Amperes or Amps (A)
R = resistance, units are ohms ()
V
I
R
Example:
1. What voltage is needed to produce a 0.3 A current
through a 100  resistor?
V = IR = (0.3 A)(100 ) = 30 V
Simple Circuits: One resistor
R = 10 
I = V/R = (10 V)/(10 ) = 1 A
V = 10 V
Circuits with more than one resistor in SERIES
10 
20 
The current has to go through
both resistors so the two
resistances add together
Rtot = 10  + 20  = 30 
V=9V
I = V/Rtot = (9 V)/(30 ) = 0.3 A
How much voltage is lost at each resistor?
10 
20 
Already know that the current
through the circuit is 0.3 A so...
V = IR = (0.3 A)(10 ) = 3 V
V=9V
V = IR = (0.3 A)(20 ) = 6 V
The total voltage dropped at all of the resistors in the
circuit have to add up to the voltage of the original power
source.
25 
10 
15 
V=5V
1. What is the total resistance of the circuit?
Rtot = 10  + 25  + 15  = 50 
2. What is the current through the circuit?
I = V/Rtot = (5 V)/(50 ) = 0.1 A
3. What is the voltage drop at each of the resistors?
V = IR = (0.1 A)(25 ) = 2.5 V
V = IR = (0.1 A)(10 ) = 1 V
V = IR = (0.1 A)(15 ) = 1.5 V
Vtot = 5 V
What does the current do if the resistors are in PARALLEL?
10 
5
The voltage source ‘sees’ the two
resistors as being just ONE. The
voltage drop across both of them is the
same.
The current through the 10  branch of
the circuit is:
I = V/R = (10 V)/(10 ) = 1 A
V = 10 V
The current through the 5  branch of the circuit is:
I = V/R = (10 V)/(5 ) = 2 A
The total current through the circuit is 3 A.
How do the resistors combine in a Parallel circuit?
10 
5
V = 10 V
1
1 1
 
Rtot R1 R2
1
1
1


Rtot 10  5 
1
 0.1  0.2  0.3
Rtot
Rtot = 3.3 
Just to check: I = V/R = (10 V)/(3.3 ) = 3 A
Power
Like with work, electrical power is the amount of electrical
work done per second.
The units of power are Watts (W)
It can be calculated from current and voltage using:
P = IV
Or it can be calculated from voltage and resistance using:
P=
2
V
R
1. How much power is dissipated in an appliance using 9 V
and 0.1 A of current?
P = IV = (0.1 A)(9 V) = 0.9 W
2. How much power is dissipated in a 100  resistor with a
12 V voltage drop across it?
P = (V2)/R = (12 V)2/(100 ) = 144/100 = 1.44 W
3. 150 W of power is dissipated by a resistor with a 15 V
voltage drop across it. What is the current through the
resistor?
I = P/V = (150 W)/(15 V) = 10 A