DC Circuits tutorial question solutions - Physics at PMB

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Oct 7, 2013 (4 years and 1 month ago)

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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 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
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 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
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