PHYSICS 6B
Ch.18 Worksheet
1) A cell phone has a battery that supplies it with a current of 0.22A. How
many electrons pass through the circuit in 4.5 s?
Q
, so we can find the total charge:
t
Q  (0.22A)(4.5s)  0.99C
0.99C
Now use electron charge:
 6.2  1018 electrons
19
C
1.6  10
Current is defined as I 
electron
2) A battery with an emf of 1.5 V delivers a current of 0.44 A to a flashlight
bulb for 64 sec. Find (a) the charge that passes through the circuit (in
Coulombs) and (b) the work done by the battery (in Joules).
Q  (0.44A)(64s)  28.2C
Power=IV=(0.44A)(1.5V)=0.66 Watts
using Power=Work/Time: W=(0.66W)(64s)=42.2J
3) Wire 1 has a length L and circular cross section of diameter D. Wire 2 is
constructed from the same material as wire 1 and has the same shape, but
its length is 2L, and its diameter is 2D. The resistance of wire 2 is
(a) half that of wire 1.
(b) the same as that of wire 1.
(c) twice that of wire 1.
(d) four times that of wire 1.
L
A
Wire 2 has twice the length, and twice the diameter (so Area is 4times
bigger than wire 1). This combines to give a resistance that is ½ as big for
wire 2. Answer a).
R 
4) A current of 1.8 A flows through a copper wire 1.75 m long and 1.1 mm
in diameter. Find the potential difference between the ends of the wire. The
resistivity of copper is 1.2 x10-8 Ω-m.
L (1.2  10 8 )(1.75m)
R 

 0.07
A
1.110 3 m 2
(
)
2
V=IR=(1.8A)(0.07Ω)=0.12V
clas.ucsb.edu/staff/vince/
PHYSICS 6B
Ch.18 Worksheet
5) A circuit consists of three resistors connected in series to a 24.0 V
battery. The current in the circuit is 0.032 A.
Given that R1=250 Ω, R2=150 Ω, find (a) the value of R3 and (b) the
potential difference across each resistor.
We can find the total resistance from Ohm’s Law: V=IR
24V
R 
 750
0.32A
Since the resistors are all in series, the total is just the sum:
750  R1  R 2  R 3  R 3  350
Now use Ohm’s law for each resistor–each gets the total current.
V1  (0.32A)(250)  8V; V2  4.8V; V3  11.2V
Note that the voltages add up to 24V, as they must.
6) Consider a circuit with three resistors, R1= 250 Ω, R2=150 Ω, and
R3=350 Ω, connected in parallel with a 24 V battery. Find (a) the total
current supplied by the battery and (b) the current through each resistor?
Total resistance is found from the reciprocal formula:
1
1
1
1



 R eq  73.9
R eq 250 150 350
24V
 0.32A
73.9
Use Ohm’s law for each individual resistor: the voltage is 24V for each:
24V
I1 
 0.096A; I2  0.16A; I3  0.069A
250
Itotal 
7) Two identical light bulbs are connected to a battery, either in series or in
parallel. Are the bulbs in series (a) brighter, (b) dimmer or (c) the same
brightness as the bulbs in parallel?
In series, the resistance is greater, so current is reduced. This means the
bulbs use less power. Thus they are dimmer when in series.
8) In a circuit two resistors are connected in series to one resistor in
parallel. The emf of the battery is 12.0V, and each resistor has a resistance
of 200 Ω. Find the current supplied by the battery to this circuit, and find the
current passing through each resistor. Also find the total power supplied by
the battery, and the power consumed by each resistor.
200Ω
200Ω
200Ω
clas.ucsb.edu/staff/vince/
The resistor network is pictured at left. It has equivalent
resistance 133.3Ω. (Combine the 2 in series, then that is in
parallel with the other path).
Total current is found from Ohm’s law: I=12/133=0.09A
The path with the single resistor has current I=24/200=0.06A
This leaves 0.03A for the upper path.
P=IV=I2R=V2/R use whichever form is convenient
Ptotal=1.08W; P=0.18W for each series resistor, P=0.72W for
the single resistor. Total adds up, as it should.