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PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
DC CIRCUITS
Question 61
A 1000 Ω 2.0 W resistor is needed but only 1000 Ω 1.0 W resistors are available. How can the
required resistance and power rating be obtained by a combinationof the available units?
What power is then dissipated in each resistor?
The current through the resistor can be found by
P = V I = I 2R =
V2
R
so the current through the 2.0 W resistor would be
r
r
1
P
2.0
=
= √ A
I=
R
1000
10 5
The maximum current that can flow through the 1.0 W resistor is
r
1.0
1
I=
= √ A
1000
10 10
hence the 1.0 W resistors must be wired in parallel. For a maximum current of 101√5 A in the
circuit, two resistors in parallel each would carry 12 101√5 < 10√1 10 A, so this is permissible.
Two 1000 Ω resistors in parallel would result in an equivalent resistance of
REQ =
1000 × 1000
= 500 Ω
1000 + 1000
so two combinations like this in series results in an equivalent resistance of 1000 Ω. The
power dissipated in each resistor is
2
1
2
√
× 1000 = 0.50 W
P =I R=
20 5
Question 62
Each of the three resistors in the figure has a resistance of 2.0 Ω and can dissipate a maximum of 18 W without becoming excessively heated. What maximum power can the circuit
dissipate?
page 1 of 6
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
The maximum current that can flow in the circuit is
r
r
P
18
=
= 3.0 A
I=
R
2.0
since this is the current flowing in the single 2.0 Ω resistor. The equivalent resistance for the
whole circuit is
2.0 × 2.0
REQ =
+ 2.0 = 3.0 Ω
2.0 + 2.0
Hence the maximum power dissipated is
P = I 2 R = 3.02 × 3.0 = 27 W
Question 63
(a) Find the potential difference Vad in the circuit
(b) Find the potential difference across the 4.00 V cell.
(c) A battery of emf 17.0 V and internal resistance 1.00 Ω is inserted in the circuit at d
with its positive terminal connected to the positive terminal of the 8.00 V battery. Find
Vbc between the terminals of the 4.00 V battery.
(a) Choosing a direction for the current as anti-clockwise and a direction for the loop as
dcbad, Kirchhoff’s second rule gives
−9I − 0.5I + 4 − 6I − 8I + 8 − 0.5I = 0
This leads to a current of
12
= 0.50 A
24
To find Vad , start at point a and move in the direction of the current to point d:
I=
Va − 8I + 8 − 0.5I = Vd
The potential difference Vad = Vd − Va is then
Vad = Vd − Va = 8 − (8 × 0.5) − (0.5 × 0.5) = 3.75 V
(b) The potential difference Vcb is found by starting at point c and moving to point d:
Vc − 0.5I + 4 = Vb
∴ Vcb = 4 − 0.25 = 3.75 V
page 2 of 6
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(c) With a new battery inserted in the circuit, the Kirchhoff II equation becomes
−9I − 0.5I + 4 − 6I − 8I + 8 − 0.5I − 17 − (1 × I) = 0
This now leads to a current of
I=−
5
= −0.20 A
25
The potential difference Vcb is then
Vcb = 4 − (0.5 × I) = 4 − (0.5 × (−0.20)) = 4.10 V
Question 64
A dry cell having an emf of 1.55 V and an internal resistance of 8.00×10−2 Ω supplies current
to a 2.00 Ω resistor.
(a) Determine the current in the circuit.
(b) Calculate the terminal voltage of the cell.
(a) The current in the circuit is given by
1.55 = (2.00 + 0.08) × I
∴I=
1.55
= 0.745 A
2.08
(b) The terminal voltage is the voltage drop across the 2.00 Ω resistor:
terminal voltage = 2.00 × 0.745 = 1.49 V
Question 65
How many cells, each having an emf of 1.5 V and an internal resistance of 0.50 Ω must be
connected in series to supply a current of 53 A to operate an instrument having a resistance
of 6.0 Ω?
page 3 of 6
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Suppose there are n cells, then the current in the circuit, the total emf and the total resistance
is given by
1.5n = (0.5n + 6.0) ×
6
× 1.5 − 0.5)
6
=
0.9 − 0.5
= 15
∴n =
5
3
( 53
Question 66
A battery has an internal resistance of 0.50 Ω. A number of identical light bulbs, each with
a resistance of 15 Ω, are connected in parallel across the battery terminals. The terminal
voltage of the battery is observed to be one-half the emf of the battery. How many bulbs
are connected?
Suppose there are n light bulbs. The total resistance for n light bulbs in parallel is
1
1
1
n
1
+
+
+ ... =
=
REQ
15 15 15
15
The terminal voltage V is
∴ REQ =
1
V = IREQ = E
2
where E = (REQ + 0.50)I. Hence
IREQ =
1
∴ REQ =
2
REQ =
⇒n =
=
1
(REQ + 0.50)I
2
1
(0.5)
2
0.5
15
0.5
30
page 4 of 6
15
n
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 68
Find the magnitude and direction of the current in the 2.0 Ω resistor in the diagram.
Using Kirchhoff’s rules, the equations needed to find the currents are:
KI at a
KII for loop abcda
KII for loop aefba
(I) gives
substitute (IV) into (II)
:
:
:
:
:
I1 + I2 + I3 = 0
−2I2 − 1 − 4 + 3I3 = 0
−I1 + 1 + 2I2 = 0
I3 = −I1 − I2
−2I2 − 5 + 3(−I1 − I2 ) = 0
−3I1 − 5I2 − 5 = 0
multiply (III) by 3 : −3I1 + 6I2 + 3 = 0
(V) - (VI) : −11I2 − 8 = 0
8
∴
I2 = − A
11
(I)
(II)
(III)
(IV)
(V)
(VI)
Hence I2 = −0.73 A is the current through the 2.0 Ω resistor and it flows from b to a.
Question 69
Determine the voltage across the 5.0 Ω resistor in the circuit below. Which end of the resistor
is at the higher potential?
The Kirchhoff equations are
KI at a : I1 + I2 = I3
KII for loop abcda : 10I2 − 15 + 2 + 10I3 = 0
KII for loop adefa : −10I3 − 2 + 10 − 5I1 = 0
page 5 of 6
(I)
(II)
(III)
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
These are solved simultaneously as follows:
rearrange (I) : I2 = I3 − I1
substitute (IV) into (II) : 10(I3 − I1 ) − 13 + 10I3 = 0
20I3 − 10I1 − 13 = 0
multiply (III) by 2 : −20I3 − 10I1 + 16 = 0
(V) + (VI) : −20I1 + 3 = 0
3
A
∴
I1 =
20
(IV)
(V)
(VI)
(VII)
The voltage drop across the 5.0 Ω resistor is
V = I1 R =
3
3
× 5.0 = = 0.75 V
20
4
Current I1 is positive, therefore it flows in the direction of f to a. That means that point a
is at a lower potential than point f.
page 6 of 6