MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 13: ELECTRIC DC
CURRENT and RESISTANCE WORKSHEET SOLUTIONS
1. The heating coil of a hot water heater has a resistance of 50.0 ohm and operates at 120 V. What is its
power rating?
1A.
(1) P = V2 / R
(2) P = (120 V)2 / 50.0 
(3) P = 288 W
2. If a certain resistor obeys Ohm’s law, its resistance will change:
(A) as voltage only across the resistor changes
(B) as current only through the resistor changes
(C) as both voltage and current change
(D) as energy given off by the electrons in their collisions changes
(E) none of the above
2A. (E) none of the above
3. A certain material is in a room at 27.0oC. If the absolute temperature K of the material is tripled, its
resistance also triples. What is the value for , the temperature coefficient of resistivity?
3A.
(1) R = Ro(1 + (T – To))
(2) (T – To) = (R/Ro) – 1
(3) R = 3Ro
(4) (T – To) = (3Ro/Ro) – 1
(5) (T – To) = 2
(6)  = 2 / (T – To)
(7)  = 2 / (900 K – 300 K)
(8)  = 2 / 600 K
(9)  = 0.0033 / oC
4. A platinum wire is utilized to determine the melting point of indium. The resistance of the platinum
wire is 2.00  at 20.0oC and increases to 3.072  as the indium starts to melt. platinum = 3.92 x 10-3/oC.
What is the melting temperature of indium (in oC)?
4A.
(1) R = Ro(1 + (T – To))
(2) T = [(R – Ro) / (Ro)] + To
(3) T = [(3.072  – 2.00 ) / (3.92x10-3/oC)(2.00 )] + 293.15 K
(4) T = 136.7 K + 293.15 K
(5) T = 429.9 K
(6) Tc = T – 273.15
(7) Tc = 156.7oC
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
5. A color television set draws 3.15 A when connected to 220 V. What is the cost (with electrical
energy at 7 cents/kWh) of running the color TV for 8 hours?
5A.
(1) Determine joules used: P = IV
(2) P = (3.15 A)(220 V)
(3) P = 693 W = 693 J/s
(4) (8 hours) (3600 sec / 1 hour) (693 J / 1 sec) = 2.00 x 107 J
(5) 2.00 x 107 J (1 kWh) = 5.54 kWh
-------------3.6 x 106 J
(6) Determine cost: (5.54 kWh) (7 cents)
(7) Cost = 38.8 cents
6. An electric car runs off a bank of 18.0 V batteries with total energy storage of 4.55 x 109 J. If the
electric motor draws 6000 W in moving the car at a steady speed of 20 m/s, how far will the car go (in
km) before it runs out of fuel?
6A.
(1) Determine time car can move:
Time = (4.55 x 107 J) (6000 W)
(2) Time = (4.55 x 107 J) (1 sec / 6000 J)
(3) Time = 7583 s
(4) Determine distance car can move:
(7583 s) (20 m/s) = 1.52 x 105 m
(5) 1.52 x 105 m (1 km / 1000 m) = 152 km
7. How long does it take for 0.2 C to pass through a frictionless wire in a 9 amp circuit?
7A.
(1) I = Q / t
(2) t = Q / I
(3) t = 0.2 C / 9A
(4) t = 2.22 x 10-2 s
8. A metal wire has a radius of 0.45 mm, is 0.85 m long, and has a resistance of 15.0 ohms. What must
be the length of a second wire made of the same type of metal if its radius is 0.10 mm and also has a
resistance of 15.0 ohms?
8A.
(1) R1 = L1 / A
(2)  = R1A / L1
(3) A = r2
(4)  = R1r2 / L1
(5)  = (15 ) (4.5 x 10-4 m)2 / 0.85 m
(6)  = 1.12 x 10-5 .m
(7) R2 = L2 / A
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
8A. (continued…)
(8) L2 = R2A / 
(9) L2 = R2(r2) / 
(10) L2 = (15 )  (1 x 10-4 m)2 / 1.123 x 10-5 .m
(11) L2 = 4.20 x 10-2 m
(12) L2 = 4.20 cm
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 13: ELECTRIC DC
DIRECT CURRENT
WORKSHEET SOLUTIONS
1. The following three appliances are connected to a 120 volt house circuit:
(i) toaster, 1350 W
(ii) coffee pot, 750 W
(iii) microwave, 600 W
If all were operated at the same time, what total current would they draw?
1A.
(1) P = IV
(2) I = P / V
(3) I = 2850 W / 120 V
(4) I = (2850 J/s) / (120 C/J)
(5) I = 23.8 C/s = 23.8 A
2. A circuit contains a 3.00 volt battery, a 2.00 ohm resistor, a 0.30 microfarad capacitor, an ammeter,
and a switch, all in series. What will be the current reading 10 min after the switch is closed?
2A. zero
3. Three resistors, with values of 3.00, 6.00, and 12.0 ohms, respectively, are connected in parallel.
What is the equivalent resistance of this combination?
3A.
(1) 1/Rp = (1/Ri)
(2) 1/Rp = 1 / 3 + 1 / 6 + 1 / 12
(3) 1/Rp = 7 / 12
(4) Rp = 12 / 7 
(5) Rp = 1.7 
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
4.
Which resistor above is in series with resistor R?
(a) R1 (b) R2 (c ) R3 (d) R4
4A. (d) R4
5. What is the equivalent resistance for these resistors?
5A.
(1) Req (top part): Req = R1 + R2
(2) Req = 6  + 10  = 16 
(3) Req (top + bottom): 1 / Req = 1 / R1 + 1 / R2
(4) 1 / Req = 1 / 16 + 1 / 4
(5) 1 / Req = 5 / 16
(6) Req = 16/5 
(7) Req (total): Req = R1 + R2
(8) Req = 2  + 16/5  = 26/5 
(9) Req = 5.2 

6. Consider the circuit below.
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
6a. Only two resistors in this circuit can be combined into one equivalent resistance. Which two are
they?
A. 2 and 1
6b. What is the value of the unspecified resistance on top?
A.
(1) Req = R1 + R2
(2) Req = 2 + 1
(3) Req = 3

6c. The next step in the circuit reduction is as shown below. What is the value of the unspecified
resistance?
A.
(1)
(2)
(3)
(4)
(5)
6d.
A.
(1)
(2)
(3)
1 / Req = 1 / R1 + 1 / R2
1 / Req = 1 / 3 + 1 / 6
1 / Req = 2 / 6 + 1 / 6
1 / Req = 3 / 6 = 1 / 2
Req = 2
How much is the single resistance representing all four of the original resistors?
Req = R1 + R2
Req = 3 + 2
Req = 5
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MR. SURRETTE
6e.
A.
(1)
(2)
(3)
(4)
(5)
6f.
A.
(1)
(2)
(3)
VAN NUYS HIGH SCHOOL
The current through the battery (and the 3 resistor) is thus:
V=IR
V = I Req
I = V / Req
I = 20V / 5
I=4A
The total power dissipated by the circuit is:
P = IV
P = (4A)(20V)
P = 80 W
7. A 20 volt battery is connected across a 5 ohm resistor, as shown.
7a. Which end of the resistor is at the higher potential, A or B?
A. Point A
7b. The 5 ohm resistor consists of a block of material with terminals at A and B. The electric field
inside this material must either be directed from A to B or from B to A. Which is it?
A.
From A to B. Electric field flows out of the positive end of the battery and into the negative end of the
battery.
7c. The direction of conventional current (electrons) through R must be:
A. It is directed from B to A through the resistor.
7d. The voltage across the 5 ohm resistor is:
A. 20V. The battery supplies the voltage.
7e.
A.
(1)
(2)
(3)
(4)
The current through the 5 ohm resistor is given by Ohm’s law and is:
V = IR
I=V/R
I = 20V / 5
I=4A
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
7f. Electrons scatter off atoms in the resistor and transfer energy (heat) to them. The rate of heat
production, or the power dissipated in the 5 ohm resistor, is:
A.
(1) P = IV
(2) P = (4A)(20V)
(3) P = 80 W
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