Download Chapter 26 Part 1-

Chapter 26 Part 1--Examples
1
Problem

If a ohmmeter is
placed between
points a and b in the
circuits to right, what
will it read?
2
Circuit a





The 75 and 40 W
resistors are in
parallel
The 25 and 50 W
resistors are in
parallel
R75-40=26.09 W
R50-25=16.67 W
Their total is 42.76 W
3
Circuit a—Cont’d

The circled network
is in parallel with the
50 W resistor so their
combined resistance
is 23.05 W.
 This resistor is in
parallel with the
original 100 W
resistor so the total
resistance is 18.7 W
4
Circuit b





The 60 and 20 W
resistor are in
parallel
The 20 W is in series
with the 30 and 40 W
parallel network.
R30-40 =18.0 W
R20-30-40= 38.0 W
R38-60=23.3 W
5
Circuit b—cont’d
The 23.03 W
equivalent network is
in series to the 7 W
resistor
 This equivalent
30.03 W resistor is
parallel to the 10 W
resistor so
 Req=7.5 W

6
Problem

In the circuit shown,
1.
2.
What must be the
EMF of the battery in
order for a current of
2 A to flow through
the 5 V battery?
How long does it take
for 60 J of thermal
energy to be
produced in the 10 W
resistor?
7
Step 1 Reduce the Resistors





10 + 20 =30
(1/30)+1/60 +1/60
=4/60 so Req=15
1/15+1/30=3/30 so
RT=10
10+5+5=20
So 20 W in the
upper network
8
Step 2 Reduce the EMF

-5 + 10=+5 V
 So the circuit
becomes:
20 W
2A
5V
15 W
EMF
20 W
9
Using our loop rules







-(2)*20-5-20I1=0
I1=-2.25 A
2=-2.25+I2
I2=4.25
-EMF-4.25*15+20*(2.25)
EMF=-108.75 V
Need to reverse the
battery….
20 W
2A
5V
15 W
EMF
I2
20 W
I1
10
2 Amps into the 10 W resistor





Since the equivalent resistance in the upper
network is 10 W and 2 A runs through it, there
is a potential difference of 20 V across each of
the legs
10+20=30 W so the current is 20/30 A=2/3 A
P=i2r so 4/9*10=40/9=4.444 W or J/s
60=4.444 * t
t =13.5 s
11