Transcript chapter 19

RESISTORS IN SERIES
- In a series circuit, the current is the same
at all points along the wire.
IT = I1 = I2 = I3
- An equivalent resistance is the resistance of a single
resistor that could replace all the resistors in a circuit.
The single resistor would have the same current through
it as the resistors it replaced.
RE= R1 + R2 + R3
- In a series circuit, the sum of the voltage drops equal
the voltage drop across the entire circuit.
VT = V 1 + V 2 + V3
RESISTORS IN SERIES
19.1 Two resistances of 2 Ω and 4 Ω respectively are connected in
series. If the source of emf maintains a constant potential
difference of 12 V,
a. What is the current delivered to the external circuit?
Re = R1 + R2
=2+4
=6Ω
V 12
IT 

=2A
Re 6
b. What is the potential drop across each resistor?
V1 = I R 1
= 2(2)
=4V
V2 = I R 2
= 2(4)
=8V
PARALLEL CIRCUITS
- In a parallel circuit, each resistor provides a new path
for electrons to flow. The total current is the sum of the
currents through each resistor.
IT = I1 + I2 + I3
- The equivalent resistance of a parallel circuit
decreases as each new resistor is added.
1
1
1
1



RE R1 R2 R3
- The voltage drop across each branch is equal to the
voltage of the source.
V T = V1 = V2 = V 3
RESISTORS IN PARALLEL
19.2 The total applied voltage to the circuit in the figure is 12 V and
the resistances R1, R2 and R3 are 4, 3 and 6 Ω respectively.
a. Determine the equivalent resistance of the circuit.
R2 and R3 are in parallel (RP)
1 1 1
 
Rp 6 3
R p= 2 Ω
RP and R1 are in series
Req = 4 + 2
=6Ω
b. What is the current through each resistor?
V 12

IT 
=2A
Re 6
I1 = 2 A (series)
V1 = IR1
= 2(4)
=8V
The voltage through the parallel combination is
therefore: 12-8 = 4 V each
V 4
I2 
 = 1.33 A
R2 3
V
4
= 0.67 A
I3 

R3
6
19.3 Find the equivalent resistance of the circuit shown.
19 and 5 are in series:
19 + 5 = 24 Ω
1 1 1
 
this combination is in parallel with 8:
Rp 8 24
RP = 6 Ω
this combination is in series with 15:
15 + 6 = 21 Ω
1
1 1
 
this in turn is in parallel with 9:
Rp 21 9
RP = 6.3 Ω
finally the equivalent resistance:
Req = 6.3 + 2 + 0.2 = 8.5 Ω
19.4 A potential difference of 20 V is applied to the circuit in the
figure below. Find the current through the entire circuit and the
current through each resistor.
1
1 1
 
R1 and R2 are in parallel:
Rp 10 10
RP = 5 Ω
this combination is in series with R3:
5+3= 8Ω
this combination is now in parallel with R4:
1 1 1
 
Rp 8 12
RP = 4.8 Ω = Req
V
20
IT 
= 4.17 A

Req
4.8
IT = I3 + I4
V
20
I4 
= 1.67 A

R4 12
I3 = IT - I4
= 4.17 - 1.67
= 2.5 A
The voltage for the parallel combination is:
V' = V - I3R3
= 20 - (2.5)(3)
= 12.5 V
V ' 12.5

I1 
= 1.25 A
10
R1
I2 = 2.5 - 1.25 = 1.25 A
EMF AND TERMINAL POTENTIAL DIFFERENCE
Every source of emf (ξ ) has an inherent resistance
called internal resistance represented by the symbol r.
This resistance is a small resistance in series with the
source of emf. The actual terminal voltage VT across a
source of emf with an internal resistance is given by:
VT = ξ - I r
Units: Volts (V)
19.5 A load resistance of 8 Ω is connected to a battery whose
internal resistance is 0.2 Ω
a. If the emf of the battery is 12 V, what current is delivered to the
load?
ξ = 12 V
RL = 8 Ω
r = 0.2 Ω
12
V


I 
= 1.46 A
8  0.2
R RL  r
b. What is the terminal voltage of the battery?
VT = ξ - I r
= 12 - 1.46(0.2)
= 11.7 V
19.6 a. Determine the total current delivered by the source of emf
to the circuit in the figure. Voltage = 24 V. The resistances are 6, 3,
1 and 2 Ω respectively.
R1 and R2 are in parallel:
1 1 1
 
Rp 6 3
RP = 2 Ω
this combination is in series with R3:
2+1= 3Ω
this combination is now in parallel with R4:
1 1 1
 
Rp 2 3
RP = 1.2 Ω
finally the internal resistance is in series giving the
equivalent resistance:
Req = 1.2 + 0.4
= 1.6 Ω
V
24 = 15 A
IT 

Req
1.6
b. What is the current through each resistor?
VT = ξ - I r
= 24 - 1.5(0.4)
= 18 V
V 18

I4 
=9A
R4 2
V4 = VT = 18 V
I3 = IT - I4
= 15 - 9
=6A
V3 = I3R3
= 6(1)
=6V
V1 = V2 = 18 - 6 = 12 V each
V 12

I1 
=2A
6
R1
V 12
=4A

I2 
3
R2
KIRCHHOFF’S LAWS
An electrical network is a complex circuit consisting of
current loops. Kirchhoff developed a method to solve
this problems using two laws.
Law 1. The sum of the currents entering a junction is
equal to the sum of the currents leaving that junction.
I in  I out
Law 2. The sum of the emfs around any closed current
loop is equal to the sum of all the IR drops around that
loop. (Ohm’s Law: V = IR)
  IR
Gustav Robert Kirchhoff
(1824-1887)
A junction refers to any point in the circuit where two or
three wires come together.
APPLYING KIRCHHOFF’S LAW
We will use the following circuit to explain the procedure
suggested by Kirchhoff:
1. Assume a current direction for each loop in the
network:
I1 counterclockwise
I2 counterclockwise
I3 counterclockwise
2. Apply Kirchhoff's first law to write a current equation
for all but one of the junction points. The equation for
junction m is:
Σ Ientering = Σ Ileaving
I1 + I2 = I3
3. Indicate by a small arrow the direction in which each
emf acting alone would cause a positive charge to move.
ξ 1 and ξ 2 are directed to the left
and ξ 3 is directed to the right.
4. Apply Kirchhoff's second law for one loop at a time.
There will be one equation for each loop. One must
begin at a specific point on a loop and trace around the
loop in a consistent direction. The tracing direction is
always positive.
The following sign conventions apply:
a. When adding emfs, the value assigned to the emf is
positive if its output is with the tracing direction. The
emf is negative if its output is against the tracing
direction.
b. An IR drop is positive when the assumed current is
with the tracing direction and negative when the
assumed current is against the tracing direction.
Loop 1 Starting at point m and tracing clockwise, we
have:
- ξ 1 + ξ 2 = - I1R1 + I2R2
Loop 2 Starting at point m and tracing counterclockwise,
we have:
ξ 3 + ξ 2 = I3R3 + I2R2
Loop 3 Starting at m and tracing counterclockwise we
have:
ξ 3 + ξ 1 = I3R3 + I1R1
We now have three independent equations involving
only three unknowns, and the third loop equation can be
used to check the results.
In order to solve the equations you may use your
calculator with MATRIX mode.
19.7 Solve for the unknown currents using Kirchhoff's laws.
+
+
FIRST LAW:
Σ Ientering = Σ Ileaving
I2 = I1 - I3
SECOND LAW:
Σ V = Σ IR
left loop:
6 V + 2V = I1 (1 Ω) + I2 (3 Ω)
8 V = 1 Ω I1 + 3 Ω I3
dividing by 1 Ω:
8 A = I1 + I3
SECOND LAW:
Σ V = Σ IR
right loop:
- 3 V = I3 (2 Ω) + I3 (4 Ω) + I2 (3 Ω)
dividing by 1 Ω:
- 3 A = 6 I3 + 3 I2
Three equations:
I1 - I2 + I3 = 0
I1 + 3 I2
=8A
I2 + 2 I3 = - 1 A
I1 = 3 A
I2 = 1.67 A
I3 = - 1.33 A
our assumed current
direction is incorrect,
therefore: I3 = 1.33 A
CHECK:
Using the outside loop counterclockwise
(6 + 2 + 3) V = 1 Ω I1 - 4 Ω I3 - 2 Ω I3
11 A = I1 - 6 I3
substituting our results:
11 A = 3 A - 6 (- 1.33 A)
11 A = 11 A
19.8 Solve for the unknown
currents using Kirchhoff's laws.
FIRST LAW:
Σ Ientering = Σ Ileaving
I2 = I1 + I3
left loop:
-20 A = 8 I2 + 10 I1
simplifying:
5 I1 + 4 I2 = -10
SECOND LAW:
Σ V = Σ IR
right loop:
- 4 = 6 I3 + 8 I2
simplifying:
4 I2 + 3 I3 = - 2
Three equations:
I1 - I2 + I3 = 0
5 I1 + 4 I2
= -10
4 I2 + 3 I3 = - 2
I1 = - 1.32 A
I2 = - 0.85 A
I3 = 0.47 A
our assumed current direction for I1 and I2 is
incorrect.
CHECK:
Using the outside loop clockwise
20 - 4 = - 10 I1 + 6 I3
substituting our results:
16 = - 10(-1.32) + 6 (0.47)
16 A = 16.02 A