PHYSICS 120:ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
DC CIRCUITS
Question 61
A 1000 Ω 2.0 Wresistor is needed but only 1000 Ω 1.0 Wresistors 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
2
R =
V
2
R
so the current through the 2.0 Wresistor would be
I =
P
R
=
2.0
1000
=
1
10
√
5
A
The maximum current that can ﬂow through the 1.0 Wresistor is
I =
1.0
1000
=
1
10
√
10
A
hence the 1.0 Wresistors must be wired in parallel.For a maximum current of
1
10
√
5
A in the
circuit,two resistors in parallel each would carry
1
2
1
10
√
5
<
1
10
√
10
A,so this is permissible.
Two 1000 Ω resistors in parallel would result in an equivalent resistance of
R
EQ
=
1000 ×1000
1000 +1000
= 500 Ω
so two combinations like this in series results in an equivalent resistance of 1000 Ω.The
power dissipated in each resistor is
P = I
2
R =
1
20
√
5
2
×1000 = 0.50 W
Question 62
Each of the three resistors in the ﬁgure has a resistance of 2.0 Ω and can dissipate a maxi
mum 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
I =
P
R
=
18
2.0
= 3.0 A
since this is the current ﬂowing in the single 2.0 Ω resistor.The equivalent resistance for the
whole circuit is
R
EQ
=
2.0 ×2.0
2.0 +2.0
+2.0 = 3.0 Ω
Hence the maximum power dissipated is
P = I
2
R = 3.0
2
×3.0 = 27 W
Question 63
(a) Find the potential diﬀerence V
ad
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
V
bc
between the terminals of the 4.00 V battery.
(a) Choosing a direction for the current as anticlockwise 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
I =
12
24
= 0.50 A
To ﬁnd V
ad
,start at point a and move in the direction of the current to point d:
V
a
−8I +8 −0.5I = V
d
The potential diﬀerence V
ad
= V
d
−V
a
is then
V
ad
= V
d
−V
a
= 8 −(8 ×0.5) −(0.5 ×0.5) = 3.75 V
(b) The potential diﬀerence V
cb
is found by starting at point c and moving to point d:
V
c
−0.5I +4 = V
b
∴ V
cb
= 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
25
= −0.20 A
The potential diﬀerence V
cb
is then
V
cb
= 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
2.08
= 0.745 A
(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
5
3
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) ×
5
3
∴ n =
6
(
3
5
×1.5 −0.5)
=
6
0.9 −0.5
= 15
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 onehalf 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
R
EQ
=
1
15
+
1
15
+
1
15
+...=
n
15
∴ R
EQ
=
15
n
The terminal voltage V is
V = IR
EQ
=
1
2
E
where E = (R
EQ
+0.50)I.Hence
IR
EQ
=
1
2
(R
EQ
+0.50)I
∴
1
2
R
EQ
=
1
2
(0.5)
R
EQ
= 0.5
⇒n =
15
0.5
= 30
page 4 of 6
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:I
1
+I
2
+I
3
= 0 (I)
KII for loop abcda:−2I
2
−1 −4 +3I
3
= 0 (II)
KII for loop aefba:−I
1
+1 +2I
2
= 0 (III)
(I) gives:I
3
= −I
1
−I
2
(IV)
substitute (IV) into (II):−2I
2
−5 +3(−I
1
−I
2
) = 0
−3I
1
−5I
2
−5 = 0 (V)
multiply (III) by 3:−3I
1
+6I
2
+3 = 0 (VI)
(V)  (VI):−11I
2
−8 = 0
∴ I
2
= −
8
11
A
Hence I
2
= −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:I
1
+I
2
= I
3
(I)
KII for loop abcda:10I
2
−15 +2 +10I
3
= 0 (II)
KII for loop adefa:−10I
3
−2 +10 −5I
1
= 0 (III)
page 5 of 6
PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS
These are solved simultaneously as follows:
rearrange (I):I
2
= I
3
−I
1
(IV)
substitute (IV) into (II):10(I
3
−I
1
) −13 +10I
3
= 0
20I
3
−10I
1
−13 = 0 (V)
multiply (III) by 2:−20I
3
−10I
1
+16 = 0 (VI)
(V) + (VI):−20I
1
+3 = 0
∴ I
1
=
3
20
A (VII)
The voltage drop across the 5.0 Ω resistor is
V = I
1
R =
3
20
×5.0 =
3
4
= 0.75 V
Current I
1
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
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