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Electric Circuits Model Worksheet 4:
More Quantitative Electric Circuit Analysis
1. Suppose the switch in the circuit at right is
closed.
6V
a. How does closing the switch affect the ∆V
from a to b?
S
c
a
e
6
12 6
630
 
No effect.
g
b
d
f
b. How does closing the switch affect the current in the first two resistors?
No effect.
c. How does closing the switch affect the current through point g?
(I)Current Before = 1.5 A (Check the resistance)
(I)Current After = 1.7 A (Check the resistance)
d. Find the total resistance of the circuit before and after the switch is closed.
Rtotal before = 4
Rtotal after = 3.5 
2. A 10 ohm and a 15 ohm resistor are connected in parallel and placed across the terminals of a
15 volt battery.
a. Sketch the circuit diagram.
10
15
15V
b. Calculate the equivalent resistance of the circuit.
1
Req
 R11  R12  101  151 
3
30 
 302 
5
30 
, R eq 
30 
5
 6.0
c. Calculate the current through the battery.
I
V 15V

 2.50A
R 6
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d. Calculate the current in each branch.
V 15V
V 15V
I10   
 1.5A, I15   
 1.0A
R 10Ω
R 15Ω
3.
Consider the circuit below. All the bulbs have resistance R.
a. What is the resistance of the circuit while the switch is
open?
Two bulbs is series is a resistance of 2R.
b. What is the resistance of the circuit when the switch is
closed?
B and C in parallel add to a resistance of ½R. A added to this gives a resistance of 1½R.
c. How does closing the switch affect the brightness of bulbs A and B? Explain in terms of
current and potential drop.
A will get brighter and B will get very dim. (C will also be very dim.)
If the resistance for the circuit decreases when the switch is closed, the current will increase. A
gets all of this current and it has a greater potential drop so it will get brighter. B will get half
of the current A gets and it has a smaller potential drop than before so it is dimmer.
To verify, call the potential difference of the battery V, before the switch is closed both A and B
have a potential difference of ½V and a current of V/2R or ½I. The power for each bulb is
½I(½V) or ¼IV. After closing the switch the circuit will have a current V/1.5R or 2/3I. The
potential drop across A is 2/3V and across B is 1/3V. The power for A is 2/3I(2/3V) or 4/9IV.
This is much greater than the ¼IV from before. The power for B is 1/3I(1/3V) or 1/9IV. As can
be seen, the power for A is 4 times that for B and that the power for B has decreased greatly
from when the switch was open.
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4. The electrical device whose symbol is shown at right requires a terminal voltage of
10 V and a current of 0.5 A for proper operation.
Using only this device and one or more 7- resistors, design a circuit so that the device will operate
properly when the circuit is connected across a 24 V battery with negligible internal resistance.
A
24V
battery
B
5. A cabin contains only two small electrical appliances: a radio that requires 10 milliamperes of
current at 9.0 V, and a clock that requires 20 milliamperes at 15 V. A 15 V battery with negligible
internal resistance supplies the electrical energy to operate the radio and clock.
a. Complete the diagram below to show how the radio,
clock and a single resistor R can be connected between
points A and B so that the correct potential difference is
applied across each appliance. Use the symbols at right
to indicate the clock and radio.
radio
10mA, 9V
clock
20 mA, 15 V
b. Calculate the resistance of R.
Rr 
V
6V

 600Ω
I .010A
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c. Calculate the electrical energy that must be supplied by the battery to operate the circuit for
one minute.
One method is to first solve for the circuit resistance. The
radio branch is 15V/.01A = 1500. The clock branch is
15V/.02A = 750. Solving for equivalent resistance
yields 500. Power for the circuit would be V2/R which
is 152/500 =.45W. Energy is power times time, so
.45Wx60s=27 joules.
6. Consider the series-parallel combination circuit at right.
6V
a. Determine the equivalent resistance of the circuit.
1
R eq
a
c
12 
 R11  R12  121  61  121  122  123 , R eq  123  4.0
6
4.0  8  12
e
8
b
d
b. What is the current passing through each resistor?
I8  
V12 
V
6V

 .50A, V8   IR  .50A  8  4V
R 12
V
2V
V 2V
 6V - 4V  2.0V, I12   
 .167A, I6   
 .333A
R 12Ω
R 6Ω
c. What is the ∆V across each resistor?
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6.0V, 2.0V, 2.0V (see work in b.)
d. What is the power dissipated by each resistor?
P  IV  .50A (4V)  2.0W, .167A (2V)  .333W, .333A(2V)  .667W
©Modeling Workshop Project 2007
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