Chapter 28A - Direct Current Circuits
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

Objectives: After completing this module, you should be able to:
• Determine the effective resistance for a number of resistors connected in series and in parallel .
• For simple and complex circuits, determine the voltage and current for each resistor.
• Apply Kirchoff’s laws to find currents and voltages in complex circuits.

Electrical Circuit Symbols
Electrical circuits often contain one or more resistors grouped together and attached to an energy source, such as a battery.
The following symbols are often used:
Ground Battery
+ -
Resistor
+ - + -
- + - + -

Resistances in Series
Resistors are said to be connected in series when there is a single path for the current.
I
V
T
R
1
R
3
R
2
The current I each resistor is the same for
R
1
, R
2 and R
3
.
The energy gained through E is lost through R
1
, R
2 and R
3
.
Only one current The same is true for voltages:
For series connections:
I = I
1
V
T
= V
1
= I
2
+ V
2
= I
3
+ V
3

Equivalent Resistance: Series
The equivalent resistance R e of a number of resistors connected in series is equal to the sum of the individual resistances.
I
V
T
R
1
R
3
R
2
V
T
= V
1
+ V
2
+ V
3
; (V = IR)
I
T
R e
= I
1
R
1
+ I
2
R
2
+ I
3
R
3
But . . . I
T
= I
1
= I
2
= I
3
Equivalent Resistance
R e
= R
1
+ R
2
+ R
3

Example 1: Find the equivalent resistance R e
What is the current I in the circuit?
.
R e
= R
1
+ R
2
+ R
3
2 W
3 W 1 W
12 V
R e
= 3 W + 2
Equivalent
W
R e
+ 1 W
= 6 W
= 6 W
The current is found from Ohm’s law: V = IR e
I

V

R e
12 V
6
W
I = 2 A

Example 1 (Cont.): Show that the voltage drops across the three resistors totals the 12-V emf.
R e
= 6 W I = 2 A
2 W
3 W 1
12 V
W
Current I = 2 A same in each R.
V
1
= IR
1
; V
2
= IR
2;
V
3
= IR
3
V
1
= (2 A)(1 W) = 2 V
V
1
= (2 A)(2 W) = 4 V
V
1
= (2 A)(3 W) = 6 V
V
1
+ V
2
+ V
3
= V
T
2 V + 4 V + 6 V = 12 V
Check !

Sources of EMF in Series
The output direction from a source of emf is from + side: a
+
E b
Thus, from a to b the potential increases by E ;
From b to a , the potential decreases by E .
Example: Find D V for path AB and then for path BA .
R
-
A
9 V
AB: D V = +9 V – 3 V = +6 V
-
3 V
+
BA: D V = +3 V - 9 V = -6 V B

A Single Complete Circuit
Consider the simple series circuit drawn below:
D
C
2 W
4 W
-
3 V
+
-
A
15 V
B
Path ABCD: Energy and V increase through the 15-V source and decrease through the 3-V source.
 E =
15 V - 3 V = 12 V
The net gain in potential is lost through the two resistors: these voltage drops are IR
2 and so that the sum is zero for the entire loop .
IR
4
,

Finding I in a Simple Circuit.
Example 2: Find the current I in the circuit below:
D
C
2 W
3 W
-
3 V
+
-
A
18 V
B
 E =
18 V 3 V

15 V

R
= 3
+ 2
W 
5
W
I
Applying Ohm’s law:



E
R

15 V
5
W
I = 3 A
In general for a single loop circuit:
I

 E

R

Summary: Single Loop Circuits:
Resistance Rule: R e
=  R
Current : I

 E

R
Voltage Rule:  E =  IR
R
2
R
1
E
1
E
2

Complex Circuits
A complex circuit is one containing more than a single loop and different current paths.
R
3
I
3
At junctions m and n: m
R
1
E
2
I
1
= I
2
+ I
3 or I
2
+ I
3
= I
1
R
2
I
1
E
1 Junction Rule:
 I (enter) =  I (leaving)
I
2 n

Parallel Connections
Resistors are said to be connected in parallel when there is more than one path for current.
Parallel Connection:
For Parallel Resistors:
2 W 4 W 6 W
V
2
I
2
= V
4
= V
6
+ I
4
+ I
6
= V
= I
T
T
Series Connection:
2 W 4 W 6 W
For Series Resistors:
I
2
= I
4
= I
6
= I
T
V
2
+ V
4
+ V
6
= V
T

Equivalent Resistance: Parallel
V
T
I
T
= V
1
= V
2
= I
1
+ I
2
= V
3
+ I
3
Ohm’s law:
I

V
R
V
T

V
1

V
2

V
R R R R e 1 2 3
3
V
T
Parallel Connection:
R
1
R
2
R
3
1

1

1

1
R R R R e 1 2 3
The equivalent resistance for Parallel resistors:
1
R e
 i
N 

1
R i
1

Example 3. Find the equivalent resistance
R e for the three resistors below.
1
R e
 i
N 

1
R i
1 V
T
2 W
R
1
4 W
R
2
R
3
6 W
1

1

1

1
R R R R e 1 2 3
1

1

1

1

R e
2
W
4
W
6
W
1
R e

0.917; R e

1
0.917

1.09
W R e
= 1.09 W
For parallel resistors, R e is less than the least R i
.

Example 3 (Cont.): Assume a 12-V emf is connected to the circuit as shown. What is the total current leaving the source of emf?
V
T
12 V
2 W
R
1
4 W
R
2
R
3
6 W
V
V
T
I
1
= 12 V; R e
T
= V
= I
2
1
= V
3
+ I
2
= 1.09 W
= 12 V
+ I
3
Ohm’s Law:
I

V
R
I e

V
T

R e
12 V
1.09
W
Total current: I
T
= 11.0 A

Example 3 (Cont.): Show that the current leaving the source I
T is the sum of the
R
1
, R
2
, and R
3
.
I
1

V
T
12 V
2 W
R
1
4 W
R
2
12 V
2
W

6 A
R
3
6 W
I
2

12 V
4
W

I
T
= 11 A; R e
V
1
= V
2
= V
3
= 1.09 W
= 12 V
I
T
= I
1
+ I
2
+ I
3
3 A I

3
12 V
6
W

2 A
6 A + 3 A + 2 A = 11 A Check !

Short Cut: Two Parallel Resistors
The equivalent resistance R e for two parallel resistors is the product divided by the sum .
1

1

1
R e
R
1
R
2
; R e

R R
1 2
R
1

R
2
V
T
Example:
R
1
R
2
6 W 3 W
R e

3 6
R e
= 2 W

Series and Parallel Combinations
In complex circuits resistors are often connected in both series and parallel .
R
1
In such cases, it’s best to use rules for series and parallel resistances to reduce the circuit to a simple circuit containing one source of emf and one equivalent resistance.
V
T
V
T
R
2
R
3
R e

Example 4.
Find the equivalent resistance for the circuit drawn below (assume V
T
= 12 V).
4 W
R
3,6

3 6
V
T
3 W 6 W
R e
= 4 W + 2 W
R e
= 6 W
4 W
12 V 2 W 12 V 6 W

Example 3 (Cont.) Find the total current I
T
.
R e
= 6 W
4 W
V
T
3 W 6 W
I

V
T

R e
12 V
6
W
I
T
= 2.00 A
4 W
12 V 2 W 12 V
I
T 6 W

Example 3 (Cont.) Find the currents and the voltages across each resistor .
4 W
I
4
= I
T
= 2 A
V
T
3 W 6 W V
4
= (2 A)(4 W ) = 8 V
The remainder of the voltage: (12 V – 8 V = 4 V ) drops across EACH of the parallel resistors.
V
3
= V
6
= 4 V
This can also be found from
V
3,6
= I
3,6
R
3,6
= (2 A)(2 W )
(Continued . . .)

V
Example 3 (Cont.) Find the currents and voltages across each resistor .
4
= 8 V V
6
= V
3
= 4 V 4 W
V
T
3 W 6 W
I
3

V
3

R
3
4 V
3
W
I
6

V
6

R
6
4 V
6
W
I
3
= 1.33 A
I
6
= 0.667 A I
4
= 2 A
Note that the junction rule is satisfied:
 I (enter) =  I (leaving) I
T
= I
4
= I
3
+ I
6

Kirchoff’s Laws for DC Circuits
Kirchoff’s first law: The sum of the currents entering a junction is equal to the sum of the currents leaving that junction.
Junction Rule:  I (enter) =  I (leaving)
Kirchoff’s second law: The sum of the emf’s around any closed loop must equal the sum of the IR drops around that same loop.
Voltage Rule:  E =  IR

Sign Conventions for Emf’s
 When applying Kirchoff’s laws you must assume a consistent, positive tracing direction.
 When applying the voltage rule , emf’s are positive if normal output direction of the emf is with the assumed tracing direction.
 If tracing from A to B , this emf is considered positive .
A
E
+
B
 If tracing from B to A , this emf is considered negative .
A
+
E
B

Signs of IR Drops in Circuits
 When applying the voltage rule , IR drops are positive if the assumed current direction is with the assumed tracing direction.
 If tracing from A to B , this
IR drop is positive .
A
I
+
B
 If tracing from B to A , this
IR drop is negative .
A
+
I
B

Kirchoff’s Laws: Loop I
1. Assume possible consistent flow of currents.
2. Indicate positive output directions for emf’s.
3. Indicate consistent tracing direction. (clockwise)
R
1
Junction Rule: I
2
= I
1
+ I
3 R
3
Voltage Rule:  E =  IR
E
1
+
E
2
= I
1
R
1
+ I
2
R
2
+
I
1
Loop I
E
2
R
2
I
3
I
2
E
1
E
3

Kirchoff’s Laws: Loop II
4. Voltage rule for Loop II:
Assume counterclockwise positive tracing direction.
Voltage Rule:  E =  IR
E
2
+
E
3
= I
2
R
2
+ I
3
R
3
Would the same equation apply if traced clockwise ?
Yes!
E
2
-
E
3
= -I
2
R
2
- I
3
R
3
R
1
R
3
Bottom Loop (II)
I
1
Loop I
E
2
R
2
I
2
I
3
Loop II
+
E
1
E
3

Kirchoff’s laws: Loop III
5. Voltage rule for Loop III:
Assume counterclockwise positive tracing direction.
Voltage Rule:  E =  IR
E
3
– E
1
= -I
1
R
1
+ I
3
R
3
Would the same equation apply if traced clockwise ?
Yes!
E
3
-
E
1
= I
1
R
1
- I
3
R
3
R
1
R
3
I
1
Loop I
E
2
R
2
I
2
I
3
Loop II
+
E
1
E
3

Four Independent Equations
6. Thus, we now have four independent equations from Kirchoff’s laws:
R
1
I
1
Loop I
E
2
R
2
I
2
= I
1
+ I
3
E
1
+
E
2
= I
1
R
1
+ I
2
R
2
E
2
+
E
3
= I
2
R
2
+ I
3
R
3
E
3
-
E
1
= -I
1
R
1
+ I
3
R
3
R
3
I
2
I
3
Loop II
+
E
1
E
3

Example 4.
Use Kirchoff’s laws to find the currents in the circuit drawn to the right.
Junction Rule: I
2
+ I
3
= I
1
+
Consider Loop I tracing clockwise to obtain:
Voltage Rule:  E =  IR
I
1
5 W
Loop I
12 V
10 W
12 V = (5 W ) I
1
+ (10 W ) I
2
Recalling that V/ W = A , gives
I
3
I
2
20 W
5 I
1
+ 10 I
2
= 12 A
6 V

Example 5 (Cont.) Finding the currents.
Consider Loop II tracing clockwise to obtain:
Voltage Rule:  E =  IR
6 V = (20 W ) I
3
- (10 W ) I
2
Simplifying: Divide by 2 and V/ W = A , gives
I
1
I
3
5 W
10 W
12 V
I
20 W
10 I
3
5 I
2
= 3 A +
6 V

Example 5 (Cont.) Three independent equations can be solved for I
1
, I
2
, and I
3
.
(1) I
2
+ I
3
= I
1
(2) 5 I
1
(3) 10 I
+ 10 I
3
5 I
2
2
= 12
= 3 A
A
I
1
5 W
10 W
12 V
Substitute Eq.
(1) for I
1
5( I
2
+ I
3
) + 10 I
3 in (2) :
= 12 A
Simplifying gives:
5 I
2
+ 15 I
3
= 12 A
+
I
3
I
20 W
6 V

Example 5 (Cont.) Three independent equations can be solved.
(1) I
2
+ I
3
= I
1
(3) 10 I
3
5 I
2
= 3 A
(2) 5 I
1
+ 10 I
2
= 12 A 15 I
3
+ 5 I
2
= 12 A
Eliminate I
2 by adding equations above right:
10 I
3
15 I
3
5 I
2
+ 5 I
2
= 3 A
= 12 A
25 I
3
= 15 A
I
3
= 0.600 A
Putting I
3
= 0.6 A in (3) gives:
10(0.6 A) – 5 I
2
I
2
= 0.600 A
= 3 A
Then from (1): I
1
= 1.20 A

Summary of Formulas:
Rules for a simple, single loop circuit containing a source of emf and resistors.
Resistance Rule: R e
Current : I

=  R
 E

R
Voltage Rule:  E =  IR
D
C
Single Loop
2 W
3 W
-
3 V
+
-
A
18 V
B

Summary (Cont.)
For resistors connected in series:
For series connections:
I = I
1
V
T
= V
1
= I
2
+ V
2
= I
3
+ V
3
R e
= R
1
+ R
2
+ R
3
R e
=  R
2 W
3 W 1 W
12 V

Summary (Cont.)
Resistors connected in parallel:
For parallel connections:
V = V
1
I
T
= I
1
= V
2
+ I
2
= V
3
+ I
3
1
R e
 i
N 

1
R i
1
R e

R R
1 2
R
1

R
2
V
T
Parallel Connection
R
1
R
2
R
3
12 V
2 W 4 W 6 W

Summary Kirchoff’s Laws
Kirchoff’s first law: The sum of the currents entering a junction is equal to the sum of the currents leaving that junction.
Junction Rule:  I (enter) =  I (leaving)
Kirchoff’s second law: The sum of the emf’s around any closed loop must equal the sum of the IR drops around that same loop.
Voltage Rule:  E =  IR

CONCLUSION: Chapter 28A
Direct Current Circuits