Chapter 11 Electricity Exercise
EXERCISES
1. 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 $R^{\prime}$, then the ratio $R / R^{\prime}$ is -
(a) $1 / 25$
(b) $1 / 5$
(c) $5$
(d) $25$
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Answer
(d) 25
2. Which of the following terms does not represent electrical power in a circuit?
(a) $I^{2} R$
(b) $I R^{2}$
(c) $V I$
(d) $V^{2} / R$
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Answer
(b) $I R^{2}$
3. 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) $25 W$
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Answer
(d) $25 W$
4. 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) $4: 1$
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Answer
(c) $1: 4$
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
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Answer
To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.
6. A copper wire has diameter $0.5 mm$ and resistivity of $1.6 \times 10^{-8} \Omega m$. What will be the length of this wire to make its resistance $10 \Omega$ ? How much does the resistance change if the diameter is doubled?
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Answer
Area of cross-section of the wire, $A=\pi(d / 2) 2$
Diameter $=0.5 mm=0.0005 m$
Resistance, $R=10 \Omega$
We know that $R=\rho_A^{l}$
$l=\dfrac{R A}{\rho}$
$=\dfrac{10 \times 3.14 \times(\dfrac{0.0005}{2})^{2}}{1.6 \times 10^{-8}}$
$=\dfrac{10 \times 3.14 \times 25}{4 \times 1.6}=122.72 m$
$\therefore$ length of the wire $=122.72 m$
If the diameter of the wire is doubled, new diameter $=2 \times 0.5=1 mm=0.001 m$ Let new resistance be $R^{\prime}$
$ \begin{aligned} & R^{\prime}=\rho_A^{l} \\ & =\dfrac{1.6 \times 10^{-8} \times 122.72}{\pi(\dfrac{1}{2} \times 10^{-3})^{2}} \\ & =\dfrac{1.6 \times 10^{-8} \times 122.72 \times 4}{3.14 \times 10^{-6}} \\ & =250.2 \times 10^{-2}=2.5 \Omega \end{aligned} $
Therefore, the length of the wire is $122.7 m$ and the new resistance is $2.5 \Omega$.
7. 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.
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Answer
The plot between voltage and current is called $I V$ characteristic. The voltage is plotted on $x$-axis and current is plotted on $y$-axis. The values of the current for different values of the voltage are shown in the given table.
| $V$ (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
|---|---|---|---|---|---|
| $I$ (amperes ) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
The IV characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance $(R)$ as,
Slope $=1 / R=B C / A C=2 / 6.8$
$R=6.8 / 2=3.4 \Omega$
Therefore, the resistance of the resistor is $3.4 \Omega$.
8. 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.
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Answer
Resistance $(R)$ of a resistor is given by Ohm’s law as, $V=I R$
$R=V / I$
Where,
Potential difference, $V=12 V$
Current in the circuit, $l=2.5 mA=2.5 \times 10^{-3} A$
$ R=\dfrac{12}{2.5 \times 10^{-3}}=4.8 \times 10^{3} \Omega=4.8 k \Omega $
Therefore, the resistance of the resistor is $4.8 k \Omega$
9. A battery of $9 V$ is connected in series with resistors of $0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega$ and $12 \Omega$, respectively. How much current would flow through the $12 \Omega$ resistor?
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Answer
There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
$V=I R$
$I=V / R$
Where,
$R$ is the equivalent resistance of resistances $0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega$ and $12 \Omega$. These are connected in series. Hence, the sum of the resistances will give the value of $R$.
$R=0.2+0.3+0.4+0.5+12=13.4 \Omega$
Potential difference, $V=9 V$
$l=9 / 13.4=0.671 A$
Therefore, the current that would flow through the $12 \Omega$ resistor is $0.671 A$.
10. How many $176 \Omega$ resistors (in parallel) are required to carry $5 A$ on a $220 V$ line?
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Answer
For $x$ number of resistors of resistance $176 \Omega$, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as $V=I R$
$R=V / I$
Where,
Supply voltage, $V=220 V$
Current, $I=5 A$
Equivalent resistance of the combination $=R$, given as
$ \begin{aligned} & \dfrac{1}{R}=x \times(\dfrac{1}{176}) \\ & R=\dfrac{176}{x} \end{aligned} $
From Ohm’s law,
$ \begin{aligned} & \dfrac{V}{I}=\dfrac{176}{x} \\ & x=\dfrac{176 \times I}{V}=\dfrac{176 \times 5}{220}=4 \end{aligned} $
Therefore, four resistors of $176 \Omega$ are required to draw the given amount of current.
11. Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of (i) $9 \Omega$, (ii) $4 \Omega$.
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Answer
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., $6 \Omega+6 \Omega+6 \Omega=18 \Omega$, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be $6 / 2=3 \Omega$ is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel
Two $6 \Omega$ resistors are connected in parallel. Their equivalent resistance will be
$ \dfrac{1}{\dfrac{1}{6}+\dfrac{1}{6}}=\dfrac{6 \times 6}{6+6}=3 \Omega $
The third $6 \Omega$ resistor is in series with $3 \Omega$. Hence, the equivalent resistance of the circuit is $6 \Omega+$ $3 \Omega=9 \Omega$.
(b) Two resistor in series
Two $6 \Omega$ resistors are in series. Their equivalent resistance will be the sum $6+6=12 \Omega$. The third $6 \Omega$ resistor is in parallel with $12 \Omega$. Hence, equivalent resistance will be
$ \dfrac{1}{\dfrac{1}{12}+\dfrac{1}{6}}=\dfrac{12 \times 6}{12+6}=4 \Omega $
Therefore, the total resistance is $4 \Omega$.
12. 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 $5 A$ ?
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Answer
Resistance $R_1$ of the bulb is given by the expression,
Supply voltage, $V=220 V$
Maximum allowable current, $I=5 A$
Rating of an electric bulb $P=10$ watts
Because $R=V^{2} / P$
$R 1=\dfrac{(220)^{2}}{10}=4840 \Omega$
According to Ohm’s law,
$V=I R$
Let $R$ is the total resistance of the circuit for $x$ number of electric bulbs
$R=V / I$
$=\dfrac{220}{5}=44 \Omega$
Resistance of each electric bulb, $R_1=4840 \Omega$
$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\ldots . .$. upto $x$ times.
$\dfrac{1}{R}=\dfrac{1}{R_1} \times x$
$x=\dfrac{R_1}{R}=\dfrac{4840}{44}=110$
$\therefore$ Number of electric bulbs connected in parallel are 110 .
13. A hot plate of an electric oven connected to a $220 V$ line has two resistance coils A and B, each of $24 \Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
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Answer
Supply voltage, $V=220 V$
Resistance of one coil, $R=24 \Omega$
(i) Coils are used separately
According to Ohm’s law,
$V=I_1 R_1$
Where,
$I_1$ is the current flowing through the coil
$I_1=V / R_1=220 / 24=9.166 A$
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, $R_2=24 \Omega+24 \Omega=48 \Omega$
According to Ohm’s law, $V=I_2 R_2$
Where,
$I_2$ is the current flowing through the series circuit
$I_2=V / R_2=220 / 48=4.58 A$
Therefore, $4.58 A$ current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, $R_3$ is given as $=$
$ \dfrac{1}{\dfrac{1}{24}+\dfrac{1}{24}}=\dfrac{24}{2}=12 \Omega $
According to Ohm’s law,
$V=I_3 R_3$
Where,
$I_3$ is the current flowing through the circuit $I_3=V / R_3=220 / 12=18.33 A$
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
14. Compare the power used in the $2 \Omega$ resistor in each of the following circuits: (i) a $6 V$ battery in series with $1 \Omega$ and $2 \Omega$ resistors, and (ii) a $4 V$ battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.
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Answer
(i) Potential difference, $V=6 V$ $1 \Omega$ and $2 \Omega$ resistors are connected in series. Therefore, equivalent resistance of the circuit, $R=$ $1+2=3 \Omega$
According to Ohm’s law,
$V=I R$
Where,
$I$ is the current through the circuit
$1=6 / 3=2 A$
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the $2 \Omega$ resistor is 2 A. Power is given by the expression, $P=(I)^{2} R=(2)^{2} \times 2=8 W$
(ii) Potential difference, $V=4 V$
$12 \Omega$ and $2 \Omega$ resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across $2 \Omega$ resistor will be $4 V$.
Power consumed by $2 \Omega$ resistor is given by
$P=V^{2} / R=4^{2} / 2=8 W$
Therefore, the power used by $2 \Omega$ resistor is $8 W$.
15. 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 $220 V$ ?
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Answer
Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be $220 V$, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating $100 W$ is given by,Power $=$ Voltage $x$ Current
Current $=$ Power/Voltage $=60 / 220 A$
Hence, current drawn from the line $=100 / 220+60 / 220=0.727$ A
16. Which uses more energy, a $250 W$ TV set in $1 hr$, or a $1200 W$ toaster in 10 minutes?
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Answer
Energy consumed by an electrical appliance is given by the expression, $H=Pt$
Where,
Power of the appliance $=P$
Time $=t$
Energy consumed by a TV set of power $250 W$ in $1 h=250 \times 3600=9 \times 10^{5} J$
Energy consumed by a toaster of power $1200 W$ in 10 minutes $=1200 \times 600$
Energy consumed by a toaster of power $1200 W$ in 10 minutes $=1200 \times 600$
$=7.2 \times 10^{5} J$
Therefore, 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.
17. An electric heater of resistance $8 \Omega$ draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
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Answer
Rate of heat produced by a device is given by the expression for power as, $P=I^{2} R$ Where, Resistance of the electric heater, $R=8 \Omega$
Current drawn, $I=15 A$
$P=(15)^{2} \times 8=1800 J / s$
Therefore, heat is produced by the heater at the rate of $1800 J / s$.
18. 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?
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Answer
(a) The melting point and of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.
(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.
(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.
(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.
