1
DC CIRCUITS
OBJECTIVES
After studying this unit, you should be able to:
1. Define or explain the meaning of the following terms:
. direct current (DC)
. electric power dissipation
. Ohm’s law
. resistance
. internal resistance
. resistivity
2 Calculate the resistance of a conductor given resistivity and the appropriate data.
3. Perform simple circuit calculations involving Ohm’s law.
4. Calculate the equivalent resistance of series, parallel and seriesparallel resistor networks.
5. Calculate the voltage, current and power dissipation in series, parallel and seriesparallel
resistor networks.
INTRODUCTION
We begin our analysis of electric circuits by considering a particular type of circuit, called a
direct current or DC circuit. In a DC circuit, current always flows in the same direction through
the circuit. For such circuits, the voltage output of the source of emf stays constant with time as
shown in Figure 1.
Figure 1
Voltage output of a DC source
The other main source of electricity is the AC (alternating current) type. As its name implies,
the direction of the current in AC electricity is constantly changing. The most common example
of an AC source of emf is the mains electricity supply. Many of the principles of DC circuits can
also be applied to AC ones. We will be studying AC electricity in the next unit.
Time t
emf
(volts)
V
2
RESISTANCE AND OHM’S LAW
So far in our study of electric circuits we have learnt of potential difference and current, and that
the first produces the second. We have also learnt that some materials are better at carrying a flow
of current than others; or, to put it the other way around, some materials offer less of a resistance
to current flow than others.
Are these three quantities  potential difference, current and resistance  related in some simple
way? It was this task of finding such a relationship that the German physicist George Ohm set
himself in the years immediately following Volta’s invention of the battery.
In 1827, Ohm showed that the flow of current in a conductor (at constant temperature) is directly
proportional to the potential difference across the conductor and inversely proportional to the
resistance of the material carrying the current. Today, this statement is known as Ohm’s law.
Letting V represent potential difference (voltage), I for current and R for resistance, we can write
Ohm’s law mathematically as follows:
R
V
I =; IR V
=
;
I
V
R =
Ohm was a humble school teacher at the time of his discovery. Instead of bringing him fame and
fortune, however, Ohm’s law embroiled him in controversy with other scientists of the day, as a
result of which he lost his school teaching position and spent the next six years in poverty and
bitter disappointment. Eventually, recognition came. Five years before his death at the age of 52,
Ohm was made Professor at the University of Munich. When he died in 1854, a statue was raised
to him and a Munich street named in honour of the man who discovered that V = RI.
Today, the unit of resistance is the Ohm, represented by the Greek symbol omega (Ω). A material
has a resistance of 1Ω if a current of 1A flows through it when an emf of 1V is applied across it.
SAQ 1
a. A PD of 20V is applied between the ends of a conductor and a current of 2A flows. Calculate
the resistance of the conductor.
b. What PD must be connected across a 100Ω resistor to obtain a current of 10mA flowing in it?
c. A car light bulb has a resistance of 6Ω. Calculate the current, which flows through it when it
is connected across the car’s 12V battery.
RESISTIVITY
The resistance of a conductor depends on both its size and shape, and on the material from which
it is made. The resistance, R, of a regularly shaped piece of material is proportional to its length,
L, and inversely proportional to its crosssectional area, A. Mathematically, we may write
A
L
R
ρ
=
3
Where ρ is a constant of the material called the resistivity. The unit of resistivity is the Ohm
metre (Ωm).
SAQ 2
Calculate the resistance of a cylindrical piece of copper wire of length 80cm and diameter 1mm.
The resistivity of copper is 1.72 × 10
8
Ωm.
So far, we have considered resistance as being the property of a single conductor to resist current
flow. In addition, electronic components are manufactured with specific resistive properties:
these are called resistors. A typical resistor is shown in Figure 2, together with its circuit symbol.
A wide range of resistors are manufactured, with the resistance value indicated in coded form by
coloured band markings.
Figure 2
Illustration of a resistor component
RESISTORS IN SERIES
When two or more resistors are connected ‘endtoend’, as shown in Figure 3, they are said to be
connected in series. If an emf is applied across the ends of the network, a current I will flow.
Since there is only one path for current to travel, it must be the same in all parts of the circuit.
That is:
I = I
1
= I
2
= I
3
(1)
Figure 3
Resistors in series across a battery
R
Tolerance Band
Colour Code
Bands
R
1
+

R
2
+

R
3
+

I
1
I
2
I
3
V
1
V
3
V
2
V
I
4
For the current I to flow through the resistors, there must be a potential difference (voltage)
across each. These are labelled V
1
, V
2
and V
3
respectively in Figure 3, with the plus sign (+)
indicating the point of highest potential and the minus sign () indicating the point of lowest
potential. The sum of the voltages V
1
, V
2
and V
3
must equal the supply voltage, V. That is:
V = V
1
+ V
2
+ V
3
(2)
Using Ohm’s law for each resistor, we can say that:
V
1
= I
1
R
1
; V
2
= I
2
R
2
; V
3
= I
3
R
3
If Rs represents the equivalent resistance of R
1
, R
2
and R
3
, then we can write:
V = V
1
+ V
2
+ V
3
IR
s
= IR
1
+ IR
2
+ IR
3
R
s
= R
1
+ R
2
+ R
3
(3)
Thus, for resistors connected in series, the equivalent resistance is equal to the sum of the
individual resistances. This rule holds for any number of resistances connected in series.
EXAMPLE
A 30V DC power supply unit (PSU) is applied across 3 resistors of value 10Ω, 20Ω and 30Ω
connected in series. Calculate the current supplied by the power supply and the voltage across
each resistor. SOLUTION
The circuit is illustrated in Figure 4.
Let I denote the current flowing in the circuit, and let V
1
, V
2
and V
3
represent the potential
differences (PDs) across the three resistors.
Total resistance, R
s
= R
1
+ R
2
+ R
3
= 10 + 20 + 30
= 60Ω
By Ohm’s law, I = V/R
= 30/60
= 0.5A
Therefore, the current supplied by the PSU is 0.5A. Now let us find the voltages, V
1
, V
2
and V
3
,
across each resistor.
V
1
= IR
1
= 0.5 × 10
= 5V
5
V
2
= IR
2
= 0.5 × 20
= 10V
V
3
= IR
3
= 0.5 × 30
= 15V
As a check, let us see if these three voltages add up to the supply voltage,
V = 30V.
V = V
1
+ V
2
+ V
3
= 5 + 10 + 15
= 30V
Figure 4
Series resistor network
SAQ 3
a. A DC electric motor, with a resistance of 3Ω, is operated from a 24V supply. What value of
resistor must be connected in series with the motor if the current through it is to be limited to
500mA?
b. When 3 identical resistors are connected in series across a 15V DC supply, a current of 10mA
flows. Find the potential difference across each resistor and its resistance value in Ohms.
V
1
V
3
V
2
+

R
1
=10
Ω
+

R
2
=20
Ω
+

R
3
=30
Ω
+

I
30V
DC Power
Supply Unit
V
6
RESISTORS IN PARALLEL
When two or more resistors are connected ‘sidebyside’, as shown in Figure 5, they are said to
be connected in parallel. When a source of emf is applied across a parallel connection of
resistors, there is more than one path through which current can flow.
Figure 5
Resistors connected in parallel across a battery
In addition, as each individual resistor is connected across the supply, the PD across each resistor
is equal to that of the supply voltage. That is:
V = V
1
= V
2
= V
3
The total current, I, flowing into the parallel circuit divides amongst the various ‘branches’ in
inverse proportion to the individual resistance in each branch. The sum of the individual branch
currents must equal the current flowing into the parallel circuit. That is:
I = I
1
+ I
2
+ I
3
(4)
Let R
p
denote the equivalent resistance of the parallel connection. Then, using Ohm’s law and
equation (4), we can write
3
3
2
2
1
1
R
V
R
V
R
V
R
V
p
++=
But V = V
1
= V
2
= V
3
. Therefore:
321
R
V
R
V
R
V
R
V
p
++=
∴
321
1111
RRRR
p
++= (5)
Therefore, for a network of resistors connected in parallel, the reciprocal of the equivalent
V
1
R
1
+

V
2
R
2
+

V
3
R
3
+

+

V
30V
I
1
I
2
I
3
I
7
resistance is equal to the sum of the reciprocals of each of the individual resistances.
A special situation arises when there are just two resistances, R
1
and R
2
, connected in parallel, as
shown in Figure 6.
Figure 6
Special case of two resistors in parallel
Using equation (5) to obtain the equivalent resistance gives:
21
21
21
111
RR
RR
RRR
p
+
=+=
resistance of sum
resistance ofproduct
21
21
=
+
=
RR
RR
R
p
The expression provides a useful shortcut to finding the equivalent resistance of two resistors
connected in parallel. It is important to remember, however, that it does not hold true for more
than two resistors. EXAMPLE
Three resistors with resistance 5Ω, 10Ω and 30Ω are connected in parallel across a 30V battery.
Calculate:
a. The current flowing in each resistor.
b. The total current supplied by the battery.
c. The equivalent resistance of the network. SOLUTION
The circuit diagram is given in Figure 7.
a. Using Ohm’s law, I
1
= V/R
1
= 30/5
= 6A
I
2
= V/R
2
R
1
R
2
R
p
is equivalent to
2
R
1
R
2
R
1
R
p
R
+
=
8
= 30/10
= 3A
I
3
= V/R
3
= 30/30
= 1A
b. Total current, I = I
1
+ I
2
+ I
3
= 6 + 3 + 1
= 10A
c. Let R
p
be the equivalent resistance. Then:
1/R
p
= 1/R
1
+ 1/R
2
+ 1/R
3
= 1/5 + 1/10 + 1/30
= 6/30 + 3/30 + 1/30
= 10/30
R
p
= 30/10
= 3Ω
As a check, the total current supplied by the battery can be calculated using the expression:
I = V/R
= 30/3
= 10A
Figure 7
Parallel resistor network
SAQ 4
Three resistors of 10Ω, 15Ω and 30Ω are connected in parallel across a supply of V volts. The
total current drawn from the supply is 2A. Calculate the supply voltage V and the current through
each resistor.
SAQ 5
What value of resistor must be connected in parallel with a 47kΩ resistor to obtain an equivalent
resistance of 42kΩ?
R
1
= 5
Ω
+

+

V
30V
I
1
I
2
I
3
I
+

R
2
= 10
Ω
+

R
3
= 30
Ω
9
Very often resistor networks contain both series and parallel circuits. Such networks can be
analysed by dealing with each series section and each parallel section separately. This procedure
is demonstrated in the following example. EXAMPLE
For the circuit shown in Figure 8, calculate:
a. The total current supplied by the battery.
b. The PD across each resistor.
c. The current flowing in each resistor.
Figure 8
Network of series and parallel resistors SOLUTION
a. To calculate the total current supplied by the battery, we need to calculate the total equivalent
resistance of the circuit. Consider firstly the equivalent resistance, R
p
, of R
3
and R
4
in parallel.
R
p
= R
3
R
4
/(R
3
+ R
4
)
= 40 × 10/(40 + 10)
= 400/50
= 8Ω
Thus, the circuit can now be redrawn as in Figure 9.
Total resistance, R = 4 + 6 + 8
= 18Ω
From Ohm’s law, I = V/R
= 18/18
= 1A
Therefore, the current supplied by the battery is 1A.
+
R
1
= 4
Ω

+
R
2
= 6
Ω

+
R
3
= 40
Ω

R
4
= 10
Ω
V
1
V
2
V
3
18V
I
10
b. Let V
1
, V
2
and V
3
denote the PDs across R
1
, R
2
and R
p
respectively.
Again from Ohm’s law, V
1
= IR
1
= 1 × 4
= 4V
V
2
= IR
2
= 1 × 6
= 6V
V
3
= IR
p
= 1 × 8
= 8V
Since R
p
represents the effective resistance of R
3
and R
4
in parallel, we can also say that the
voltage V
3
= 8V is the voltage across R
3
and R
4
in Figure 8.
c. The current flowing in R
1
and R
2
is the battery current, I = 1A. Let I
1
and I
2
denote the
currents flowing through R
3
and R
4
respectively.
From Ohm’s law, I
1
= V
3
/R
3
= 8/40
= 0.2A
I
2
= V
3
/R
4
= 8/10
= 0.8A
As a check, determine if I = I
1
+ I
2
= 1A
I = I
1
+ I
2
= 0.2 + 0.8
= 1A
Figure 9
Equivalent circuit
+
R
1
= 4
Ω

+
R
2
= 6
Ω

V
1
V
2
V
3
18V
I
+
R
p
= 8
Ω

11
SAQ 6
For the circuit shown in Figure 10, determine the current through the 20Ω resistor.
Figure 10
Circuit for SAQ 6
KIRCHOFF’S LAWS
So far we have analysed electric circuits by simplifying them, stepbystep, into a form to which
we can apply Ohm’s law. For relatively simple circuits this procedure is quite adequate.
However, it is not always easy to reduce more complex circuits to a form in which Ohm’s law
can be directly applied. In such situations, Kirchoff’s two laws for electric networks are
extremely useful.
First, let us define two terms. A junction is a point in a circuit at which three or more conducting
paths are joined. A loop is any closed conducting path in an electric circuit.
We can now state the two laws of Gustav Kirchoff (18241887), the eminent German physicist,
developed by him when he was only twenty years old.
Kirchoff’s first law (the current law) states that:
The total current flowing into any junction equals the total current flowing out of it.
If we adopt the convention that current flowing into a junction is positive and that current flowing
out is negative, we can restate Kirchoff’s current law as follows:
The algebraic sum of the currents at any junction in a network is zero.
Consider Figure 11.
Figure 11
Illustration of Kirchoff’s current law
R
3
= 6
Ω
R
1
= 9
Ω
R
5
= 15
Ω
R
4
= 20
Ω
R
2
= 12
Ω
40V
I
5
I
1
I
2
I
3
I
4
Junction
(node)
I
1
+ I
4
= I
2
+ I
3
+ I
5
12
The current entering junction = current leaving junction
I
1
+ I
4
= I
2
+ I
3
+ I
5
Or,
I
1
 I
2
 I
3
+ I
4
 I
5
= 0
Kirchoff’s second law (the voltage law) states that:
In any closed loop in a network, the algebraic sum of the voltage drops (PDs) around the loop is
equal to the emf acting in that loop.
Consider Figure 12.
Figure 12
Illustration of Kirchoff’s voltage law
In this case, there is only one loop.
emf = sum of voltage drops
= V
1
+ V
2
+ V
3
= IR
1
+ IR
2
+ IR
3
In applying Kirchoff’s voltage law, we must carefully assign the correct polarity (positive or
negative value) to the voltage drops. The general convention used is as follows. When
progressing around a loop, a PD going from high to low in the direction of progress is taken as
positive; conversely, a PD going from low to high is taken as negative. Therefore, if we progress
around a loop and the current is in the same direction as our progress, the PD is taken as positive.
If the current direction is opposite to our direction of progress, the PD is taken as negative.
EXAMPLE
As an example of the application of Ohm’s and Kirchoff’s laws consider the circuit shown in
Figure 13. Given the information indicated on the diagram, calculate the values of I
1
and R
2
.
SOLUTION
From Ohm’s law  : V
3
= I
3
R
3
= 3 × 4 =12V, which is also equal to V
2
since the resistors R
2
and
R
3
are in parallel.
From Kirchoff’s Voltage law  : V
1
+V
2
= 36V. Therefore V
1
= 36 – V
2
= 36  12 = 24V .
E
V
1
+

V
3
+

V
2
+

E = V
1
+ V
2
+ V
3
13
From Ohm’s law  : I
1
= V
1
/ R
1
= 24 / 6 = 4A.
From Kirchoff’s current law  : I
1
= I
2
+ I
3
. Therfore I
2
= I
1
– I
3
= 4  3 = 1A.
From Ohm’s law  : R
2
= V
2
/ I
2
= 12 / 1 = 12Ω.
Figure 13
Illustration of Ohm’s and Kirchoff’s Laws
INTERNAL RESISTANCE
So far the voltage sources in our circuits have been ideal in that they have had zero internal
resistance. Practical voltage sources, including power supplies and batteries possess some
internal resistance. When a voltage source delivers current to a circuit a potential difference is
developed across this internal resistance and so not all of the e.m.f of the source is produced at its
terminals; some potential (voltage) is lost across the internal resistance of the source. Practically,
the internal resistance of a source can be represented as a resistance, r, in series with the e.m.f. as
shown in Figure 14 (a). Now consider this source connected to an external resistance R as shown
in Figure 14(b). This circuit is just a source of e.m.f which causes a current to flow in a resistive
circuit consisting of two resistances, r and R, connected in series, which is a simple circuit and
can be analysed using the methods we have been studying. The main point is that because there is
a voltage drop across the internal reistance r, the voltage developed across the external resistance
R is less than the e.m.f. value E. Thus in circuit analysis problems where it is required to take into
account the internal resistance of the source, the technique is to place the internal resistance value
in series with the source of e.m.f and then treat it like any other resistive component.
Figure 14
Internal Resistance of a Voltage Source
ELECTRIC POWER

+
A
+

B
E
(b)
.E
rR
R
AB
V
+
=
rR
E
I
+
=
r
+

V
r
A
+

r
V
AB
= E
B
E
(a)
+

+

R
1
= 6
Ω
V
1
+

R
2
V
3
V
2
I
3
= 3A
R
3
= 4Ω
+

I
1
R
14
Suppose we have a current I flowing through a resistance R, as illustrated in Figure 15, where the
PD across the resistance is V.
Figure 15
Illustration of power dissipation in a resistor
If the current, I, flows for a time t seconds, then a charge of Q coulombs will be transferred from
B to A, where Q = It. It may be shown that (see notes on Electric Fields), the work done, W, in
transferring this charge Q from B to A, against a potential V, is given by:
W = QV = VIt joules
This work done by the voltage source is expended in the form of thermal energy (heat). The
energy dissipated per second in the resistor is defined as the power dissipation in the resistor. It
is measured in watts. Hence:
P = W/t
= VI × t/t
= VI watts
From Ohm’s law:
V = RI or I = V/R
P = VI
= (IR)I
= I
2
R
Also:
P = VI
= V(V/R)
= V
2
/R
Therefore, there are three, equivalent expressions for electric power dissipation in a resistance.
These are:
P = VI; P = I
2
R; P = V
2
/R
Heat
B A
I
+

V
R
15
EXAMPLE
For the circuit shown in Figure 16, calculate:
a. The power delivered by the battery.
b. The power dissipated in each resistor.
c. Verify that the total power dissipation is equal to the power delivered by the battery.
Figure 16
Two resistor circuit SOLUTION
a. I = 10/(3 + 2)
= 2A
The power, P, delivered by the battery is:
P = VI
= 10 × 2
= 20W
b. Let P
1
and P
2
denote the power dissipation in R
1
and R
2
respectively:
P = I
2
R
1
= (2)
2
× 3
= 12W
P = I
2
R
2
= (2)
2
× 2
= 8W
c. Total power dissipation = P
1
+ P
2
= 12 + 8
= 20W
This equals the total power delivered by the battery, as calculated in part a. of this example.
R
1
=3
Ω
R
2
= 2 Ω
I
10V
16
SAQ 7
a. A car light bulb is rated at 24W, 12V. What current flows in the bulb when it is switched on
and what is its resistance?
b. An electric soldering iron has an element of resistance 100Ω and carries a steady current of
0.5A for 1 hour. How much electrical energy is converted into heat energy during this period?
What is the power rating of the iron?
SUMMARY
1. A DC source of emf is one whose voltage output remains constant with time.
2. The resistance R of a conductor of length L and crosssectional area A is given by:
R = ρL/A where ρ is a constant of the material known as the resistivity.
3. Ohm’s law states that resistance, voltage and current are all inter related as follows:
R = V/I; V = RI; I = V/R
4. In a series resistor circuit:
Rs = R
1
+ R
2
+ R
3
+ ........ + R
n
I = I
1
= I
2
= I
3
= ............. = I
n
V = V
1
+ V
2
+ V
3
+ ........ + V
n
5. In a parallel resistor circuit:
1/R
p
= 1/R
1
+ 1/R
2
+ 1/R
3
+ ...... + 1/R
n
I = I
1
+ I
2
+ I
3
+ ................... + I
n
V = V
1
= V
2
= V
3
= ............. = V
n
In the special case of two parallel resistors:
R
p
= R
1
R
2
/(R
1
+ R
2
) (= product/sum)
6. For seriesparallel resistor networks, each series and parallel section has to be treated
separately and their equivalent resistances calculated. These are progressively substituted in
the combined circuit until the original circuit is reduced to a simple series or parallel circuit.
7. When an electric current flows in a resistance R, electrical energy from the source of emf is
converted into heat. The power dissipation, P, in watts, which defines the rate at which
energy is converted, is given by the three equivalent expressions:
P = VI; P = I
2
R; P = V
2
/R.
8. The power, P, produced by a source of emf, V, which delivers a current, I, to a circuit is given
by:
17
P = VI.
ANSWER TO SAQS
SAQ 1
a. A PD of 20V is applied between the ends of a conductor and a current of 2A flows. Calculate the
resistance of the conductor.
b. What PD must be connected across a 100
Ω
resistor to obtain a current of 10mA flowing in it?
c. A car light bulb has a resistance of 6
Ω
. Calculate the current, which flows through it when it is
connected across the car’s 12V battery.
From Ohm’s Law: R = V/I; V = IR; I = V/R
a. R = V/I
= 20/2
= 10Ω
b. V = IR
= 10 × 10
3
× 100
= 1V
c. I = V/R
= 12/6
= 2A
SAQ 2
Calculate the resistance of a cylindrical piece of copper wire of length 80cm and diameter 1mm. The
resistivity of copper is 1.72 × 10
8
Ω
m.
The crosssectional area, A, of the wire is given by the expression:
A = πr
2
= π (0.0005)
2
m
2
= 7.85 × 10
7
m
2
The resistance of the wire is given by the expression:
R = ρL/A
where: ρ = 1.72 × 10
 8
Ohmmetres
L = 0.8m
18
A = 7.85 × 10
7
m
2
Therefore:
R = (1.72 × 10
8
× 0.8)/(7.85 × 10
7
)
= 0.0175Ω
SAQ 3
a. A DC electric motor, with a resistance of 3
Ω
, is operated from a 24V supply. What value of resistor
must be connected in series with the motor if the current through it is to be limited to 500mA?
b. When 3 identical resistors are connected in series across a 15V DC supply, a current of 10mA flows.
Find the potential difference across each resistor and its resistance value in Ohms.
a. The circuit is illustrated in Figure 17
Let R = the resistance of series resistor
Rm = resistance of motor
I = V/(R + R
m
)
0.5 = 24/(R + 3)
0.5(R + 3) = 24
0.5R + 1.5 = 24
0.5R = 22.5
R = 45Ω
Figure 17
Circuit for SAQ 3a
b. The circuit is illustrated in Figure 18
Let R represent the resistance of each resistor. The total resistance, R
T
, is therefore:
R
T
= R + R + R = 3R
R
T
= V/I
∴3R = 15/(10 × 10
3
)
= 1500Ω
R = 1500/3
= 500Ω
+

V
24V
R
Motor
R = 3
Ω
I = 500mA
= 0.5A
19
V
1
= V
2
= V
3
= IR
= 10 ×10
3
× 500
= 5V
Figure 18
Circuit for SAQ 3b
SAQ 4
Three resistors of 10
Ω
, 15
Ω
and 30
Ω
are connected in parallel across a supply of V volts. The total
current drawn from the supply is 2A. Calculate the supply voltage V and the current through each resistor.
The circuit is illustrated in Figure 19
Let R
p
represent the equivalent resistance of R
1
, R
2
and R
3
in parallel.
1/R
p
= 1/ R
1
+ 1/ R
2
+ 1/ R
3
= 1/10 + 1/15 + 1/30
= 3/30 + 2/30 + 1/30 Rp = 30/6
= 5Ω
V = IR
p
= 2 × 5
= 10V
Let I
1
, I
2
and I
3
denote the currents through R
1
, R
2
and R
3
respectively.
I
1
= V/R
1
= 10/10
= 1A I
2
= V/R
2
= 10/15
= 0.67A I
1
= V/R
3
= 10/30
= 0.33A
+

V
2
+

V
3
+

V
1
I = 100 mA
V =15V
R
R
R
20
Figure 19
Circuit for SAQ 6
SAQ 5
What value of resistor must be connected in parallel with a 47k
Ω
resistor to obtain an equivalent
resistance of 42k
Ω
?
For two resistors R
1
and R
2
in parallel, their equivalent resistance R
p
is given by the expression:
R
p
= R
1
R
2
/(R
1
+ R
2
)
In the case, R
p
= 42kΩ and R
1
= 47kΩ
The problem is to calculate R
2
:
47 × R
2
/(47 + R
2
) = 42
47R
2
= 42(47 + R
2
)
47R
2
= 1974 + 42R
2
5R
2
= 1974
R
2
= 394.8kΩ
SAQ 6
For the circuit shown, determine the current through the 20
Ω
resistor.
Circuit for SAQ 6
To find the current through the 20Ω resistor R
4
, we need to know the voltage across it. Since R
2
, R
4
and R
5
are in parallel, it is possible to reduce these to a single series circuit, as shown below. It is now easy to find
R
3
= 6
Ω
R
1
= 9
Ω
R
5
= 15
Ω
R
4
= 20
Ω
R
2
= 12
Ω
40V
V
I = 2A
R1 = 10Ω
I
1
R2 = 15Ω
I
2
R3 = 30Ω
I
3
21
the voltage across R
p
and hence R
2
.
1/R
p
= 1/R
2
+ 1/R
5
+ 1/R
4
= 1/12 + 1/15 + 1/20
= 5/60 + 4/60 + 3/60
= 12/60 R
p
= 60/12
= 5Ω
Total resistance, R = R
1
+ R
p
+ R
3
= 9 + 5 + 6
= 20Ω
Current, I = V/R
= 40/20
= 2A
Voltage across R
p
= IR
p
= 2 × 5
= 10V
Current through R
4
= I
4
I
4
= 10/20
= 0.5A
Equivalent Circuit for SAQ 6
V
Rp
40V
R
3
= 6
Ω
I
R
1
= 9
Ω
22
SAQ 7
a. A car light bulb is rated at 24W, 12V. What current flows in the bulb when it is switched on and what is
its resistance?
b. An electric soldering iron has an element of resistance 100
Ω
and carries a steady current of 0.5A for 1
hour. How much electrical energy is converted into heat energy during this period? What is the power
rating of the iron?
a. Power = voltage × current
P = VI
24 = 12I
I = 2A
Resistance, R = V/I
= 12/2
= 6Ω
b. Electrical energy transferred = power × time
= (VI) × t
= (I
2
R) × t
In this case:
I = 0.5A
R = 100Ω
t = 1 hour = 60 × 60 seconds = 3600 seconds
Energy converted to heat = (0.5)
2
× 100 × 3600
= 90,000 joules
= 90kJ
Power, P = I
2
R
Power rating of solderingiron = (0.5)
2
× 100
= 25W
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