Lesson: Electricity
A
piece of wire of resistance R is cut into five equal parts. These parts are
then connected in parallel. If the equivalent resistance of this combination is
R', then the ratio $\frac{R}{R\text{'}}$ is:
(a)
$\frac{1}{25}$
(b) $\frac{1}{5}$
(c) 5
(d) 25
d
Which
of the following terms does not represent electrical power in a circuit?
(a) ${I}^{2}R$
(b) $I{R}^{2}$
(c) $VI$
(d) $\frac{{V}^{2}}{R}$
a
An
electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power
consumed will be:
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
d
Two
conducting wires of the same material and of equal lengths and equal diameters
are first connected in series and then parallel in a circuit across the same
potential difference. The ratio of heat produced in series and parallel
combinations would be:
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
d
How is a voltmeter connected in the circuit to
measure the potential difference between two points?
A voltmeter should be
connected in parallel to measure the potential difference between two points.
A
copper wire has diameter 0.5 mm and resistivity of $1.6\times {10}^{-8}\Omega \text{}m$.
What will be the length of this wire to make its resistance 10 Ω? How much does
the resistance change if the diameter is doubled?
Area
of cross-section of the wire, $A=\pi {\left(\frac{D}{2}\right)}^{2}$.
Radius $=\frac{\text{diameter}}{2}=\left(\frac{0.5}{2}\right)\text{\hspace{0.17em}}\text{mm}=\left(\frac{0.0005}{2}\right)\text{\hspace{0.17em}}\text{m}$
Resistance, $R=10\text{}\Omega $.
The resistance is given by,
$R=\rho \frac{l}{A}$. Therefore the length of the wire,
$l=\frac{RA}{\rho}$
$=\frac{10\times 3.14\times \frac{{(0.0005)}^{2}}{{2}^{2}}}{1.6\times {10}^{-8}}$
$=\frac{10\times 3.14\times 25}{1.6\times 4}$
$=\text{}122.72\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{m}$
If the diameter of the wire is doubled,
the new diameter $=2\times 0.5=1\text{mm}=0.001\text{m}$
Therefore, the new resistance
$=\rho \frac{l}{A}$
$=\frac{1.6\times {10}^{-8}\times 122.72\times 4}{\pi \times {(\frac{1}{2}\times {10}^{-3})}^{2}}$
$=\frac{1.6\times {10}^{-8}\times 122.72\times 4}{3.14\times {10}^{-5}}$
$=250.2\times {10}^{-2}$
$=25\text{\hspace{0.17em}}\Omega $
The
values of current I flowing in a given resistor for the
corresponding values of potential difference V across the
resistor are given below:
I(amperes) 0.5, 1.0,
2.0, 3.0,
4.0
V (volts) 1.6, 3.4, 6.7, 10.2,
13.2
Plot
a graph between V and I and calculate the
resistance of that resistor.
The coordinates for the graph is given
below.
V (volts) |
1.6 |
3.4 |
6.7 |
10.2 |
13.2 |
I (amperes) |
0.5 |
1.0 |
2.0 |
3.0 |
4.0 |
The
corresponding graph is given below.
Resistance of
the resistor:
The resistance of the resistor is equal to
the slope of the line formed by the V-I graph.
So, the slope $=\frac{V}{I}=\frac{BC}{AC}=\frac{6.8}{2}$
$R=\frac{6.8}{2}=3.4\text{}\Omega $.
Therefore, the resistance is 3.4 Ω.
When
a 12 V battery is
connected across an unknown resistor, there is a current of 2.5 mA in the
circuit. Find the value of the resistance of the resistor.
Resistance of a resistor, $R=\frac{V}{I}$
Here the potential difference, V = 12 V and the current
in the circuit,
$\begin{array}{l}I=2.5\text{mA}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2.5\text{}x\text{}{10}^{-3}\text{A}\end{array}$
Therefore,
$\begin{array}{l}R=\frac{1.2}{2.5\times {10}^{-3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.8\times {10}^{3\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\Omega \end{array}$
Therefore, the resistance of the resistor is $4.8\times {10}^{3}\Omega $.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistors?
Since all the resistors are connected in
series, the current flow through all the component is the same. The equivalent
resistance R
$\begin{array}{l}=0.2\text{}\Omega +0.3\text{}\Omega +0.4\text{}\Omega +0.5\text{}\Omega +12\text{}\Omega \text{}\\ =13.4\text{}\Omega .\end{array}$
As per Ohm’s law:
$\begin{array}{l}V=IR\text{\hspace{0.17em}}\\ \Rightarrow I=\frac{V}{R}\end{array}$
Potential difference, $V=\text{}9\text{V}$
$\Rightarrow I=\frac{9}{13.4}=0.671\text{A}$
As the current flow through each resistance will be the same in a series
circuit, the current that will flow through the 12 Ω resistor is 0.671 A.
How
many 176 Ω resistors (in parallel) are required to carry 5 A on a 220V line?
As per Ohm’s law V = IR
$\Rightarrow R=\frac{V}{I}$
Here, the supply voltage, $V=220$
V and the current, $I=5$
A
Let n be the number of resistors
connected in parallel.
The equivalent resistance of the
combination = R, is given as:
$\begin{array}{l}\frac{1}{R}=n\times \frac{1}{176}\\ \Rightarrow R=\frac{176}{n}\end{array}$
From Ohm’s law,
$\begin{array}{l}\frac{V}{I}=\frac{176}{n}\\ n=\frac{176\times I}{V}\\ =\frac{176\times 5}{220}\\ =4\end{array}$
Therefore, four resistors of 176 Ω are required to draw 5A of current on a 220 V line.
Show
how you would connect three resistors, each of resistance 6 Ω, so that the
combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
a)
Connecting
all the resistors in series:
The
equivalent resistance $=6\text{}\Omega +6\text{}\Omega +6\text{}\Omega =18\text{}\Omega $
b)
Connecting
all the resistors in parallel:
The
equivalent resistance will be $=2\text{}\Omega $
As both the results are not desired, a
combination of series and parallel
connections would be required.
c) Connecting two resistors in parallel:
When
two 6 Ω resistors are connected in parallel. Their equivalent resistance is:
$\begin{array}{l}=\frac{1}{\frac{1}{6}+\frac{1}{6}}\\ =\frac{6\times 6}{6+6}\\ =\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\Omega \end{array}$
Connecting
the third resistor in series with 3 Ω, we get the equivalent resistance $=6\text{}\Omega +3\text{}\Omega =9\text{}\Omega $
d) Connecting two resistors in series:
When
two 6 Ω resistors are in series, the equivalent resistance will be their sum,
$6+\text{}6=12\text{}\Omega $
Connecting
the third resistor in parallel with 12 Ω, we get the equivalent resistance as:
$\begin{array}{l}=\frac{1}{\frac{1}{12}+\frac{1}{6}}\\ =\frac{12\times 6}{12+6}\\ =\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\Omega \end{array}$
The
combination of resistors, as done in the previous two cases, give the desired
resistance.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Given:
Supply voltage, $V=220$
V
Maximum allowable current, $I=5$
A
Rating of an electric bulb, $P=10$ watt.
Let R_{1 }be the resistance
of each lamp.
As the resistance, $R=\frac{{V}^{2}}{P}$
$\Rightarrow {R}_{1}=\frac{{(220)}^{2}}{10}=4840\text{\hspace{0.17em}}\Omega $
According to Ohm’s law,
$V=IR$
Let R be the total resistance of the circuit
for n number of electric bulbs
$\Rightarrow R=\frac{V}{I}=\frac{220}{5}=44\text{\hspace{0.17em}}\Omega $
Since the resistance of each electric bulb, ${R}_{1}=4840\text{\hspace{0.17em}}\Omega $
$\begin{array}{l}\Rightarrow \frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\mathrm{.....}\text{uptontimes}\\ \Rightarrow \frac{1}{R}=\frac{1}{{R}_{1}}\times n\\ \Rightarrow n=\frac{{R}_{1}}{R}\\ =\frac{4840}{44}\\ =110\end{array}$
Therefore, the number
of electric bulbs connected in parallel are 110.
A
hot plate of an electric oven connected to a 220 V line has two resistance
coils A and B, each of 24 Ω resistances, which may be used separately, in
series, or in parallel. What are the currents in the three cases?
Given:
The supply voltage, $V=220\text{V}$
The resistance of one coil, $R=24\text{}\Omega $
Case 1: When the coils are used separately
Let I_{1} be the current
flowing through the coil.
According to Ohm's law,
$V={I}_{1}{R}_{1}$_{}
$\Rightarrow {I}_{1}=\frac{V}{{R}_{1}}=\frac{220}{24}=9.166\text{A}$
Thus,9.16 A current will flow through the
coil when used separately.
Case 2: When the coils are connected in
series
Let the total resistance of the circuit be R_{2}and the total current flowing be I_{2.}
The total resistance, ${R}_{2}=24\text{}\Omega +24\text{}\Omega =48\text{}\Omega $
According to Ohm's law, $V={I}_{2}{R}_{2}$
$\Rightarrow {I}_{2}{}_{}=\frac{V}{{R}_{2}}=\frac{220}{48}=4.58\mathrm{A}$
Thus, when the coils are connected in series, 4.58A current will flow through
the circuit.
Case 3: When the coils are connected in
parallel
Let the total resistance of the circuit be
R_{3} and the current through
it be I_{3}.
The total resistance, R_{3} is given as
$\frac{1}{\frac{1}{24}+\frac{1}{24}}=\frac{24}{2}=12\text{\hspace{0.17em}}\Omega $
According to Ohm's law,
$V={I}_{3}{R}_{3}$
$\Rightarrow {I}_{3}=\frac{V}{{R}_{3}}=\frac{220}{12}=18.33\text{A}$
Thus,
18.33 A current will flow through the circuit.
Compare
the power used in the 2 Ω resistor in each of the following circuits:
(i)
a 6 V battery in series with 1
Ω and 2 Ω resistor, and
(ii)
a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Case (i)
Given:
·
Potential difference, $V=6$ V
·
1 Ω and 2 Ω resistors are
connected in series.
The
equivalent resistance of the circuit, $R=1+2=3\text{}\Omega $
Let
I be the current through the circuit.
According to Ohm’s law,
$\begin{array}{l}V=IR\\ \Rightarrow I=\frac{6}{3}=2\text{A}\end{array}$
So,
the current flowing through the 2 Ω resistor is 2 A.
The
power used $={\left(I\right)}^{2}R={\left(2\right)}^{2}\times 2=8\text{W}$
Case (ii)
Given:
·
Potential
difference, $V=4$
V
·
12 Ω and
2 Ω resistors are connected in parallel.
Since the resistors are connected in parallel, the voltage
across 2 Ω resistor is 4 V.
The power consumed by the resistor $=\frac{{V}^{2}}{R}=\frac{{4}^{2}}{2}=8\text{W}$
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V,
are connected in parallel to electric mains supply. What current is drawn from
the line if the supply voltage is 220 V?
Since both the bulbs are connected in parallel, the potential
difference across each of them is same and is equal to 220 V.
The current drawn by the bulb of rating 60 W $=\frac{\text{power}}{\text{voltage}}=\frac{60}{220}\text{A}$.
The current drawn by the bulb of rating 100 W $=\frac{\text{power}}{\text{voltage}}=\frac{100}{220}\text{A}$
Therefore, the total current drawn $=\frac{\text{60}}{\text{220}}+\frac{100}{220}=0.727\text{\hspace{0.17em}}\text{A}$
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W
toaster in 10 minutes?
If P is the
power of the appliance and t is the
time for which it is in use,
then, the energy consumed by the electrical appliance H=
Pt.
Therefore, the energy consumed by the TV set of
power 250 W in 1 h
$\begin{array}{l}=250\times 3600\\ =9\times {10}^{5}\text{J}\end{array}$
The energy consumed by the toaster of power 1200 W in 10
minutes
$\begin{array}{l}=1200\times 600\\ =7.2\times {10}^{5}\text{J}\end{array}$
Thus, the energy consumed by a 250 W TV set in 1 h is more
than the energy consumed by a toaster of power 1200 W in 10 minutes.
An electric heater of resistance 8 Ω draws 15 A from the
service mains 2 hours. Calculate the rate at which heat is developed in the
heater.
If I is the current flowing through the circuit and R
is the resistance of the circuit, the rate of heat produced by a device is the
expression for power of the device which is equal to ${I}^{2}R$.
Given:
The resistance of the electric heater, $R=8\text{}\Omega $
The current drawn, $I=15\text{A}$
Therefore, the power, $P={\left(15\right)}^{2}\times 8=1800\text{J/s}$
Thus, the heat produced by the heater is at the rate of 1800
J/s.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament
of electric lamps?
(b) Why are the conductors of electric heating
devices, such as bread-toasters and
electric irons, made of an
alloy rather than a pure metal?
(c) Why is the series arrangement not used for
domestic circuits?
(d) How does the resistance of a wire vary with
its area of cross-section?
(e) Why are copper and aluminium wires usually
employed for electricity
transmission?
(a) Tungsten is an alloy. It has a very high melting point and
a very high resistivity. Thus, it offers very high resistance when current
flows through it. Due to the high resistance, the tungsten becomes very hot and
glows.
(b)
The heating element of the heater is made up of alloy. An alloy offers very
high resistance when current flows through it. Due to the high resistance, the
alloy becomes very hot and glows. The heating elements are not made up of pure
metal because metals are good conductor of electricity. Thus, they offer very
little resistance in flow of electric current. Very little heat is produced
when current passes through it.
(c) Series arrangements are not used in the domestic
circuits for the following reasons:
i) In a series
connection, the overall resistance of the circuit becomes high due to which the
current supply from the power source is very low.
ii) If one appliance in series connection stops
working, the other appliances too stop working. Similarly, if we have to use
only appliance, the other appliances will also start working.
iii) All the appliances connected in series connection
do not get the same voltage as supplied by power supply.
(d) When the area of cross section
increases, the resistance decreases, and vice versa. Thus, the resistance (R)
of a wire is inversely proportional to its area of cross-section (A).
(e) Copper and aluminium are good conductors of electricity.
Thus, during transmission, the loss of current would be least. That is the
reason why copper and aluminium are usually employed for electricity
transmission.