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Q:

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:

(a) Due to high melting point and high resistivity of tungsten, it does not burn at a high temperature. As the electric lamps glow at very high temperatures hence, the tungsten is used as heating element of electric bulbs.

(b) Bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of pure metals. Due to high resistivity, they produce large amount of heat.

(c) In a series circuit, there is a voltage distribution. Each element of a series circuit receives a small voltage for a large supply voltage. Thus, the amount of current decreases and the device becomes hot. Therefore, series arrangement is not used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A).

(e) The wires made of copper or aluminium have low resistivity and they are good conductors of electricity. Therefore, they are frequently used for electricity transmission.

Q:

An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

A:

$\begin{array}{l} G i v e n, R = 8 Ω \\ I = 15 A \\ R a t e of heat produced can be expressed as power, \\ {P = I}^{2} R \\ O n putting the values, \\ P = {(15 A)}^{2} \times 8 = 1800 W or 1800 \frac{J}{s} . \end{array}$

Q:

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

A:

\begin{array}{l} Energy consumed by a TV set of power 250 W in 1 h \\ is given by the expression \\ H = Pt   = 250 W \times 3600 s = 9 \times 10^{5} J \\ S i m i l a r l y, e n e r g y consumed by a toaster of power 1200 W \\ in 10 minutes  = 1200 \times 600 = 7 .2 \times 10^{5} J \\ Hence, the energy consumed by a 250 W TV in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes. \end{array}

Q:

Two lamps, one rated 100 W and 220 V, and the other 60 W at 220 W, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

A:

\begin{array}{l} T h e p o t e n t i a l difference across each bulb 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, \\ {P = VI}_{1} \\ ∴ I_{1} = \frac{P}{V} = \frac{100 W}{220 V} \\ S i m i l a r l y, current drawn by the bulb of rating 60 W is \\ given by, \\ I_{2} = \frac{P}{V} = \frac{60 W}{220 V} \\ ∴ Current drawn from the line  = \frac{100 W}{220 V} + \frac{60 W}{220 V} = 0.727 A \end{array}

Q:

Compare the power used in the 2 Ω in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

A:

$\begin{array}{l} (i) Given, Potential difference, V = 6 V \\ R_{1} {= 1 Ω, R}_{2} = 2 Ω \\ ∴ {Equivalent resistance in series, R = R}_{1} {+ R}_{2} \\ = (1+2) Ω = 3 Ω \\ On using the relation, \\ V = IR (I = current through the circuit) \\ ∴ I = \frac{V}{R} = \frac{6}{3} = 2 A \\ Now, power in the circuit \\ {P = I}^{2} {R = (2)}^{2} ×2=8 W . \\ (ii) Given, potential difference, V = 4 V \\ R_{1} = 12 Ω, R_{2} = 2 Ω \\ The voltage across each component of a \\ parallel circuit remains the same . Hence, voltage across \\ 2 Ω resistor will be 4 V . \\ Power consumed by 2 Ω is given by \\ P = \frac{V^{2}}{R} = \frac{4^{2}}{2} = 8 W . \end{array}$

Q:

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 resistances which may be used separately, in series or in parallel. What are the currents in the three cases?

A:

$\begin{array}{l} G i v e n, Supply voltage, V = 220 V \\ Resistance, R = 24 Ω \\ (i) When coils are used separately: \\ By using the relation, \\ {V = I}_{1} R_{1} \\ {Where, I}_{1} = current flowing through the coil \\ ∴ I_{1} = \frac{V}{R_{1}} = \frac{220}{24} = 9 .166 A . \\ (ii) When coils are connceted in series: \\ total resistance of the two coils \\ R_{2} = (24+24) Ω = 48 Ω \\ By using the relation, \\ {V = I}_{2} R_{2} \\ ∴ I_{2} = \frac{V}{R_{2}} = \frac{220}{48} = 4 .58 A . \\ (iii) When coils are connceted in parallel: \\ {Total resistance (R}_{3}) would be \\ \frac{1}{R_{3}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} \\ o r, R_{3} = 12 Ω \\ On using the relation \\ {V = I}_{3} R_{3} \\ ∴ I_{3} = \frac{V}{R_{3}} = \frac{220}{12} = 18.33 A . \end{array}$

Q:

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?

A:

\begin{array}{l} G i v e n, V = 220 V \\ i = 5 A \\ P_{1} = 10 W \\ On using the relation, \\ P = \frac{V^{2}}{R_{1}} \\ {or,         R}_{1} = \frac{V^{2}}{P_{1}} \\ ∴ = \frac{{(220)}^{2}}{10} = 4840 Ω \\ As per ohm's law, \\ V= IR \\ ∴ R = \frac{V}{I} = \frac{220}{5} = 44 Ω \\ Re s i s \tan c e of each electric bulb, \\ R_{1} = 4840 Ω \\ ∴ \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{1}} + .... + u p t o x times \\ or,        x = \frac{R_{1}}{R} = \frac{4840}{44} = 110 \\ T h u s, 110 electric bulbs are connected in parallel . \end{array}

Q:

Show how would you connect three resistors, each of resistance 6 , so that the combination has a resistance of (i) 9 (ii) 4.

A:

(i)

\begin{array}{l} When two resistors are connected in parallel and \\ the third one is connected in series with these \\ two then, Two 6 Ω resistors are connected in parallel . \\ Their equivalent resistance (R') will be \\ \frac{1}{R'} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \\ o r, R' = 3 Ω \\ The third 6 Ω resistor is connected in series with 3 Ω . \\ {Hence, the equivalent resistance R}_{1} = (6+3) Ω =9 Ω \end{array}

(ii)

$\begin{array}{l} Two resistors of 6 Ω are in series . Their equivalent \\ resistance will be = (6+6) Ω = 12 Ω \\ The third 6 Ω resistor is connected in parallel with 12 Ω . \\ {Hence, equivalent resistance (R}_{2}) \\ \frac{1}{R_{2}} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} = \frac{1}{4} \\ {or, R}_{2} = 4 Ω \end{array}$

Q:

How many 176 resistors (in parallel) are required to carry 5 A on a 220 V line?

A:

\begin{array}{l} G i v e n, V = 220 volt \\ I = 5 A \\ Equivalent resistance (R) of the combination is \\ given as \\ \frac{1}{R} = x \times \frac{1}{176} \\ or,       R = \frac{176}{x} \\ From ohm's law, \\ \frac{V}{I} = \frac{176}{x} \\ o r, x = \frac{176 \times 5}{220} = 4 \\ H e n c e, 4 resistors of 176 Ω are required to draw the \\ given amount of current . \end{array}

Q:

A battery of 9 V is connected in series with resistors of 0.2 , 0.3 , 0.4 , 0.5 and 12 , respectively. How much current would flow through the 12 resistor?

A:

\begin{array}{l} T h e current would be same throughout the series circuit . \\ E q u i v a l e n t resistance of the circuit, \\ R_{eq} = (0.2 + 0.3 + 0.4 + 0.5 + 12) Ω = 13.4 Ω \\ V = 9 V \\ By using ohm's law, \\ V =IR \\ o r, I = \frac{V}{R} \\ ∴ I = \frac{9}{13.4} =0 .671 A \\ Therefore, the current that would flow through \\ 12 Ω resistor is 0 .671 A . \end{array}

Q:

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.

A:

\begin{array}{l} A c c o r d i n g to ohm's law, \\ V = IR \\ or,                   R = \frac{V}{I} \\ Here, potential difference, V = 12 volts \\ i  = 2 .5 mA \\ =2 .5 \times 10^{-3} A \\ ∴ R = \frac{12}{2.5 \times 10^{-3}} =4 .8 \times 10^{-3} Ω . \end{array}

Q:

The value 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.

A:

When the graph is plotted between voltage and current then, it is called V-I characteristics. Here, the voltage is plotted on X axis and current is plotted on Y axis.

\begin{array}{l} T h e slop of the line gives the value of resistance (R) \\ as, \\ Slope, = \frac{1}{R} = \frac{B C}{A C} = \frac{2}{6.8} \\ ∴ R = \frac{6.8}{2} = 3.4 Ω . \end{array}

Q:

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ m. What will be the length of this wire to make its resistance 10 ? How much does the resistance change if the diameter is doubled?

A:

\begin{array}{l} Given, ρ = 1.6 \times 10^{- 8} Ω m \\ A = π r^{2} \\ r = \frac{0.5}{2} m m = \frac{0.0005}{2} m \\ R = 10 Ω \\ H e n c e, length of the wire \\ l = \frac{R A}{ρ} = \frac{10 \times 3.14 {(\frac{0.0005}{2})}^{2}}{1.6 \times 10^{- 8}} = 122.72 m \\ If the diameter of the wire is doubled then, new diameter \\ = 2 \times 0 .5 mm =1 mm =0 .001m \\ Hence,  Resistance R' = ρ \frac{l}{A} = \frac{1.6 \times 10^{- 8} \times 122.72}{π \times {(\frac{1}{2} \times 10^{- 3})}^{2}} \\ = 2 .5 Ω \end{array}

Q:

How is a voltmeter connected in the circuit to measure the potential difference between two points?

A:

A voltmeter need to be connected in parallel in a circuit to measure the potential difference between two points.

Q:

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

A:

\begin{array}{l} The correct option is (c) . \\ Suppose,  R = resistance of each wire \\ ∴ The equivalent resistance in series \\ R_{s} = R+R = 2R \\ If V is the applied potential difference, then it is the \\ voltage across the equivalent resistance . \\ {V = I}_{s} \times {2R         (I}_{s} = current in series combination) \\ \Rightarrow I_{s} = \frac{V}{2 R} \\ T h e heat dissipated in time t is, \\ H_{s} = I_{s}^{2} \times 2 R \times t = {(\frac{V}{2 R})}^{2} \times 2 R \times t \\ \Rightarrow H_{s} = \frac{V^{2} t}{2 R} ... (i) \\ T h e equivalent resistance of the parallel connection is \\ R_{p} = \frac{1}{\frac{1}{R} + \frac{1}{R}} = \frac{R}{2} \\ V {is the applied potential difference across  R}_{p} . \\ ∴ V = I_{p} \times \frac{R}{2} \end{array}

Q:

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

A:

\begin{array}{l} {Given,  V}_{1} {= 220 V,  V}_{2} = 110 V \\ P_{1} = 100 W \\ O n using the relation, \\ P_{1} = \frac{V_{1}^{2}}{R} \\ or,           R = \frac{V_{1}^{2}}{P_{1}} = \frac{{(220 V)}^{2}}{(100 W)} = 484 Ω \\ If the bulb is operated on 110 V, then the energy \\ consumed, \\ P_{2} = \frac{V_{2}^{2}}{R} = \frac{{(110 V)}^{2}}{484 Ω} = 25 W \\ Therefore, the power consumed will be 25 W . \end{array}

Q:

Which of the following terms does not represent electrical power in a circuit?

(a) I²R

(b) IR²

(c) VI
$(d) \frac{V^{2}}{R}$

A:

\begin{array}{l} Electrical power is given by, \\ P= VI ... (1) \\ According to ohm's law, \\ V = IR ... (2) \\ Where, V = potential difference \\ I = current \\ R = resistance \\ ∴ P = VI \\ From equation (1), P = (IR)×I \\ ∴ {P = I}^{2} R \\ From equation (2), \\ I = \frac{V}{R} \\ ∴ P = \frac{V^{2}}{R} \\ {Hence,   P = VI = I}^{2} R = \frac{V^{2}}{R} \\ {Thus, IR}^{2} does not represent electrical power . \end{array}

Q:

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 $\frac{R}{R'}$ is
$\begin{array}{l} (a) \frac{1}{25} \\ (b) \frac{1}{5} \\ (c) 5 \\ (d) 25 \end{array}$

A:

The correct option is (d).

\begin{array}{l} The resistance of a piece of wire is R and it is proportional \\ to the length of wire . As the wire is cut into five equal parts \\ h e n c e, resistance of each part = \frac{R}{5} \\ Let the equivalent resistance of these piece of wire in \\ parallel is R' then, \\ \frac{1}{R'} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} = \frac{25}{R} \\ o r, \frac{R}{R'} = 25 \end{array}

NCERT Solutions For Class 10 Science Chapter 12 Electricity

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