Solutions to Exercises
1. An
electric circuit is illustrated below.
What is the equivalent resistance of
this circuit?
R_{2} + R_{3}
= R_{eq}
6 + 6 = 12 W
1/Rp = 1/R_{1} +
1/R_{eq}
1/Rp = 1/4+ 1/12
Rp = 3 W
Rt = 3 + 6 = 9 W
2. An
electric circuit is illustrated below.
What is the value of resistor R_{3}?
I_{1} = I_{2} + I_{3}
because I1 is the total current in this case.
2 = 1.5 + I_{3}
I_{3} = 0.5 A
If V_{2} = 90 V, then V_{3}
= 90 V because voltage is constant in parallel.
R_{3} = V_{3}/I_{3}
= 90/0.5 = 180 W
3. A
seriesparallel electric circuit is illustrated below.
Find R_{t}.
R_{4} and R_{3} are in series so R_{eq1} = R_{4} + R_{3} = 10 + 20 = 30W.
R_{2 }and R_{eq1} are in parallel, so
1/ R_{eq2}
= 1/ R_{2}+ 1/ R_{eq1}
1/ R_{eq2}
= 1/ 15+ 1/ 30
R_{eq2}
= 10 W.
Rt = R_{5}
+ R_{eq2}+ R_{1 }= 10 + 10 + 10 = 30 W.
4. A
seriesparallel electric circuit is illustrated below.
What is the potential difference
across the terminal of resistor R_{1}?

First find total resistance: 1/R_{eq} = 1/R_{1}
+ 1/(R_{2 }+ R_{3}) 1/R_{eq} = 1/30+ 1/(5_{
}+ 10) 
=
10 W.
R_{t }= R_{4}
+ R_{eq } = 20 + 10 = 30 W.
_{ } I_{t }= V/R = 12/30 = 0.40 A
V_{4}
= IR_{4} = 0.40(20) = 8 V
V_{1
}= V_{t } V_{4} = 12 8 = 4 V
5. A
seriesparallel electric is illustrated below.
What is the intensity of the current
flowing from the power source, I_{s}?
R_{2 }and R_{3} experience the same voltage,
so V_{2} = I_{2}R_{2} = (0.5)75 = 37.5 V and
I=
V_{3}/R_{3} = 37.5/100 = 0.375A
So
total current = 0.5 + 0.375 = 0.875 A
6. The following electric circuit consists
of a power supply, five resistors (R_{1}, R_{2}, R_{3},
R_{4} and R_{5}) and an ammeter.
The ammeter reads 0.25 A.
a. What is the potential difference
(voltage), V_{t}, across the
terminals of the power supply?
1/ R_{eq1} = 1/(R_{1} + R_{2}) + 1/R_{3}
1/ R_{eq1}
= 1/(20 + 40) + 1/30
R_{eq1}
= 20 W.
1/ R_{eq2} = 1/R_{4} + 1/R_{5}
1/ R_{eq2}
= 1/40 + 1/120
R_{eq2} = 30 W._{}
Rt = R_{eq1}
+ R_{eq2} = 20
+ 30 = 50 W.
Vt = IRt =
0.25 (50) = 12.5 V
b. What is the potential difference across R_{3}?
Vp =V_{3 }= I_{t} R_{eq1}
=0.25 (20) = 5V
c. What
is the potential difference across R_{1}?
I_{1}
= Vp/(R_{1}+R_{2}) = 5/(20 + 40)= 0.0833A
V_{1} = I_{1}R_{1}
= 0.0833(20) = 1.67V
d. What current flows through R_{5}?
V_{5} = Vt V_{3} = 12.5  5 = 7.5 V
I_{5}
= V_{5}/R_{5} = 7.5/120 = 0.0625 A
7. An
electric circuit is illustrated below.
What is the current intensity, I, in resistors R_{2} and R_{3}?
R_{eq} = 1/R_{4} + 1/(R_{2}+R_{3})
R_{eq}
= 1/10 + 1/( 3 + 7)
R_{eq}
= 5 W.
Rt
= R_{1} + R_{eq} = 7 + 5 = 12 W.
It
= Vt/Rt = 6/12 = 0.5 A
The
current flowing through R_{2} and R_{3 }will be the same as whats on top (since R is the
same), so it will be 0.5/2 = 0.25A
8. The following electric circuit consists
of a power source, five resistors (R_{1}, R_{2}, R_{3},
R_{4} and R_{5}) and two ammeters and .
What is the potential difference
(voltage) across the terminals of resistors R_{3}?
R_{2} and R_{3} are receiving 1.5 0.75 = 0.75
A.(same current!) That implies that R_{2} + R_{3} = R_{4}.
So
R_{3} = 20 10 = 10
W.
9. The following circuit consists of a
power source, two ammeters and, a voltmeter and three resistors (R_{1},
R_{2} and R_{3}).
The total current intensity I_{t} is 20 A. Current intensity I_{3} is 12 A. The
potential difference (voltage) V_{1}
across the terminals of resistor R_{1} is 5 V.
What is the resistance of resistor R_{3}?
I_{1} = It I_{3} = 20 12 = 8 A = I_{2}
V_{2}
= I_{2}R_{2} = 8(5) = 40 V.
V_{1}
+ V_{2} = V_{3}
5
+ 40= V_{3} = 45 V
R_{3}= V_{3}/I_{3} = 45/12 = 3.75 W.
10. A source with a potential difference of
30 V is connected to the circuit shown below.
What is the current intensity I across the circuit?
V_{2} = I_{2}R_{2} = 1(10) = 10 V
V_{3}
= Vt V_{2} = 30 10 = 20 V
It
= V_{3}/R_{3} = 20/10 = 2 A
11. The following electrical
circuit consists of a power source, four resistors (R_{1}, R_{2},
R_{3} and R_{4}) and a voltmeter V_{4} (V_{s} = V_{total}).
What is the current intensity (I_{3})
through R_{3}?
V_{2and3} = 100 60 = 40 V
I_{3} = V_{2and3}/(R_{2
}+ R_{3})= 40/(10+30) = 1.0 A
12. How can one 25 W and two 100 W resistors be connected
so that their total resistance is 75 W?
Place the two 100 W resistors in parallel
(so Req = 50 W)and put the new branch in series with the 25 W.
13. How
can four 1.0 W resistors and one 2.0 W resistor be connected
to give a combined resistance of 1.5 W?
14. Four identical
resistors are connected as shown. If the total voltage is 12V, find the voltage
across each resistor.
R_{1} has the full 12V.
Its not in series with anything.
R_{2} has twice the
resistance of R_{3} and R_{4}
combined (confused? Just assign any value(for example 1 ohm to all 4 identical resistors, and
work it out), so it will have twice the voltage. 12 = 2x + x; x = 4 V.
Conclusion V_{2} = 2x = 8V;
V_{3} and V_{4} = 4V.