Electricity

Intext Questions 1

Question 1

What does an electric circuit mean?

Answer

A continuous and closed path of an electric current is called an electric circuit.

Question 2

Define the unit of current.

Answer

The unit of current is ampere (A). One ampere is constituted by the flow of one coulomb of charge per second, i.e., 1A = $\dfrac{1 \text{C}}{1 \text{s}}$

Question 3

Calculate the number of electrons constituting one coulomb of charge.

Answer

Charge on one electron (e) = 1.6 × 10^-19 C

q = 1 C

Let n electrons constitute one coulomb of charge.

∴ q = ne

where,

q = charge
e = charge on one electron
n = number of electrons

⇒ n = $\dfrac{\text{q}}{\text{e}}$

⇒ n = $\dfrac{1}{1.6 \times 10^{-19}}$

= 6.25 x 10^-18 ≈ 6 x 10^-18

Therefore, the number of electrons constituting one coulomb of charge is 6 x 10^-18

Intext Questions 2

Question 1

Name a device that helps to maintain a potential difference across a conductor.

Answer

Battery/Cell.

Question 2

What is meant by saying that the potential difference between two points is 1 V?

Answer

When 1 Joule of work is done to move a charge of 1 coulomb from one point to another, then the potential difference between the two points is said to be 1 V.

$\therefore 1 \text{ Volt} = \dfrac{1 \text{ Joule}}{1 \text{ Coulomb}}$

Question 3

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer

Given,

Potential difference (V) = 6V

Charge (q) = 1 C

We know,

V = $\dfrac{W}{q}$

Hence, Work done for 1 C will be W = Vq

Substituting we get, W = 6 x 1 = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V battery.

Intext Questions 3

Question 1

On what factors does the resistance of a conductor depend?

Answer

The resistance of a conductor depends on :

1. Its length
2. Its area of cross section
3. The nature of its material

Question 2

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer

R = ρ $\dfrac{l}{A}$

Area of the cross-section of wire is inversely proportional to the resistance, so, the thick wire offers less resistance, hence, current will flow more easily through it.

Question 3

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer

According to Ohm's Law:

I = $\dfrac{\text{V}}{\text{R}}$

Given, the potential difference decreases to half of its former value, keeping the resistance constant

V₂ = $\dfrac{\text{V}}{2}$ , R₂ = R, I₂ = ?

Substituting the values we get,

I₂ = $\dfrac{\text{V}}{2\text{R}}$ = $\dfrac{\text{I}_1}{2}$

Therefore, the current flowing through the electrical component is reduced by half.

Question 4

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer

Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices like electric iron, toasters etc.

Question 5

Use the data in Table 11.2 to answer the following :

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Answer

(a) Iron with a lesser resistivity of 10.0 x 10^-8 is a better conductor than mercury having a resistivity of 94.0 x 10^-8.

(b) Silver is the best conductor because it's resistivity of 1.60 x 10^-8 is the least in the table.

Intext Questions 4

Question 1

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Answer

Question 2

Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer

According to Ohm's Law,

V = IR

The total resistance of the circuit is 5 Ω + 8 Ω + 12 Ω = 25 Ω.

Potential difference (V) = 6 V,

I = $\dfrac{\text{V}}{\text{R}} = \dfrac{6}{25}$ = 0.24 A

Hence, current shown in ammeter = 0.24 A

Let potential difference across the 12 Ω resistor be V₁.

V₁ = 0.24 × 12 = 2.88 V

Therefore, the voltmeter reading will be 2.88 V.

Intext Questions 5

Question 1

Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω, 10³ Ω, and 10⁶ Ω.

Answer

(a) 1 Ω and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^6} = \dfrac{10^6 +1}{10^6}$

R_p = $\dfrac{10^6}{10^6 +1}$ ≈ $\dfrac{10^6}{10^6}$

R_p = 1 Ω

Therefore, the equivalent resistance is 1 Ω (approx).

(b) 1 Ω, 10³ Ω, and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^3} + \dfrac{1}{10^6} = \dfrac{10^6 + 10^3 + 1}{10^6} = \dfrac{1001001}{1000000}$

R_p = $\dfrac{1000000}{1001001}$ = 0.999 Ω

Therefore, the equivalent resistance is 0.999 Ω.

Question 2

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer

Given,

Potential difference across circuit = 220 V

Resistance of electric lamp = 100 Ω

Resistance of toaster = 50 Ω

Resistance of water filter = 500 Ω

Equivalent resistance across the three appliances connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{100}} + \dfrac{1}{\text{50}}+ \dfrac{1}{\text{500}} = \dfrac{5+10+1}{\text{500}} = \dfrac{16}{\text{500}}$

Hence, R_p = $\dfrac{500}{\text{16}} = \dfrac{125}{\text{4}} = 31.25 \space \Omega$

The resistance of the electric iron connected to the same source will be 31.25 Ω as same current flows through it as that across the three appliances (connected in parallel).

Current across electric iron = ?

Applying Ohm's law

V = IR

Substituting we get,

220 = I x 31.25

I = $\dfrac{220}{31.25}$ = 7.04 A

Therefore, current through the electric iron will be 7.04 A.

Question 3

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer

Advantages of parallel connection over series connection are:

1. Independent Operation — Devices connected in parallel operate independently of each other. Failure of one device does not affect the others. In series connection, if one device fails, the whole circuit goes down.
2. Voltage Stability — In parallel connection, each device receives the same voltage as the battery regardless of the number of devices connected. This ensures consistent device performance and prevents the voltage drop that occurs in series circuits as more devices are added.
3. Reduced Total Resistance — In parallel connection, the effective resistance of the circuit is reduced. This is helpful when each gadget has a different resistance and requires different current to operate properly.
4. Easy Troubleshooting — Troubleshooting parallel circuits is often easier because a fault in one device does not affect the others. In a series circuit, locating a fault can be more challenging because it disrupts the entire circuit.

Question 4

How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of

(a) 4 Ω,

(b) 1 Ω?

Answer

(a) If we connect 3 Ω and 6 Ω in parallel, the equivalent resistance will be less than 3 Ω. Now, if 2 Ω is connected in series with this equivalent resistance, we should get a total resistance of 4 Ω.

So, the circuit will be as shown below:

3 Ω and 6 Ω are connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{2+1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$Ω

Hence, R_p = 2 Ω

The equivalent resistor 2 Ω is in series with the 2 Ω resistor.

R_eq= 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) Since the total resistance in this case is less than the resistance of each resistor hence, the resistors 2 Ω, 3 Ω and 6 Ω must be connected in parallel to get a total resistance of 1 Ω.

So, the circuit will be as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = \dfrac{6}{6} = 1Ω$

R_eq = 1Ω

Hence, the total resistance of the circuit is 1 Ω.

Question 5

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer

(a) To secure the highest total resistance, the four coils should be connected in series as shown below:

R_eq = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) To secure the lowest total resistance, the four coils should be connected in parallel as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24} \\[1em] = \dfrac{6+3+2+1}{24} \\[1em] = \dfrac{12}{24} = \dfrac{1}{2} \Omega$

R_eq = 2Ω

Hence, the lowest total resistance is 2 Ω.

Intext Questions 6

Question 1

Why does the cord of an electric heater not glow while the heating element does?

Answer

The heating element of an electric heater is made of an alloy having high resistance and when the current flows through it, the heating element becomes too hot and glows red. The cord on the other hand is made of copper or aluminum having low resistance, so current easily flows through it producing minimal heat. Hence, it doesn't glow.

Question 2

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer

Given,

t = 1 hour = 3600 s
V = 50 V
Q = 96000 C

I = $\dfrac{\text{Q}}{\text{t}} = \dfrac{96000}{3600} = \dfrac{80}{3}$A

According to Joule's law :

Heat (H) = Voltage (V) x current (I) x time (t)

Substituting we get,

H = 50 x $\dfrac{80}{3}$ x 3600 = 4.8 x 10⁶ J

Therefore, amount of heat generated = 4.8 x 10⁶ J

Question 3

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer

Given,

R = 20 Ω
I = 5 A
t = 30 s

According to Joule's law of heating,

H = I²Rt

Substituting the values, we get,

H = 5 x 5 x 20 x 30 = 15000 J

Therefore, amount of heat developed = 15000 J.

Intext Questions 7

Question 1

What determines the rate at which energy is delivered by a current?

Answer

Power determines the rate at which energy is delivered by a current.

Question 2

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer

Given,

I = 5 A
V = 220 V
t = 2 hr = 7200 s
P = ?

We know

Power (P) = Voltage (V) x Current (I)

Substituting the values, we get

P = 220 × 5 = 1100 W

The energy consumed by the motor (E) = P × T

Substituting the values, we get

E = 1100 × 7200

= 79,20,000 J

= 7.92 × 10⁶ J

Hence, power of the motor is 1100 W and the energy consumed is 7.92 × 10⁶ J.

Exercises

Question 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', then the ratio $\dfrac{\text{R}}{\text{R}'}$ is ...............

1. $\dfrac{1}{25}$

2. $\dfrac{1}{5}$

3. 5

4. 25

Answer

25

Reason — Given,

Wire of resistance R is cut into five equal parts hence, resistance of each part is $\dfrac{\text{R}}{5}$

$\dfrac{1}{\text{R}'} = \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}}+ \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} \\[1em] = \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} \\[1em] = \dfrac{25}{\text{R}} \\[1em] \Rightarrow \text{R}' = \dfrac{\text{R}}{25}$

Hence,

$\dfrac{\text{R}}{\text{R}'} = \dfrac{\text{R}}{\dfrac{\text{R}}{25}} \\[1em] \Rightarrow \dfrac{\text{R}}{\text{R}'} = 25$

Question 2

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

1. I²R
2. IR²
3. VI
4. $\dfrac{\text{V}^2}{\text{R}}$

Answer

IR²

Reason — Electrical power in a circuit:

P = VI = I²R = $\dfrac{\text{V}^2}{\text{R}}$

Question 3

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ...............

1. 100 W
2. 75 W
3. 50 W
4. 25 W

Answer

25 W

Reason —

Given, V = 220V ; P = 100 W ; R = ?

We know,

$\phantom{\Rightarrow} \text{P} = \dfrac{\text{V}^2}{\text{R}} \\[1em] \Rightarrow \text{R} = \dfrac{\text{V}^2}{\text{P}}$

Substituting the values, we get

R = $\dfrac{\text{220}^2}{\text{100}}$ = 484 Ω

When voltage is reduced, resistance remains the same. Hence, the power consumed :

P = $\dfrac{110^2}{484}$ = 25 W

Hence, the power consumed = 25 W.

Question 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

1. 1:2
2. 2:1
3. 1:4
4. 4:1

Answer

1:4

Reason — Let R_s and R_p be the equivalent resistance of the wires when connected in series and parallel respectively.

R_s = R + R = 2R

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} = \dfrac{2}{\text{R}}$

R_p = $\dfrac{\text{R}}{2}$

Let P_s and P_p be the power consumed in series and parallel circuits, respectively.

Power (P) = $\dfrac{\text{V}^2}{\text{R}}$

P_s = $\dfrac{\text{V}^2}{\text{2R}}$

and

P_p = $\dfrac{\text{V}^2}{\dfrac{\text{R}}{2}} = \dfrac{2\text{V}^2}{\text{R}}$

For the same potential difference V, the ratio of the heat produced in the circuit is given by ratio of P_s and P_p hence

$\text{H}_s : \text{H}_p = \dfrac{\dfrac{\text{V}^2}{\text{2R}}}{\dfrac{2\text{V}^2}{\text{R}}} \\[1em] = \dfrac{\text{V}^2 \times \text{R}}{\text{2R} \times 2\text{V}^2} \\[1em] = \dfrac{1}{4}$

Hence, the ratio of the heat produced is 1:4.

Question 5

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

Answer

In the circuit, voltmeter is connected in parallel to measure the potential difference between two points.

Question 6

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^–8 Ω 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?

Answer

Given,

diameter = 0.5 mm

area = $\dfrac{πd^2}{4} = \dfrac{3.14 \times (5 \times 10^{-4})^2}{4}$ = 19.625 x 10^–8 m

resistivity = 1.6 × 10^–8 Ω m

R = 10 Ω

l = ?

We know, R = ρ $\dfrac{l}{A}$

Substituting we get,

10 = 1.6 × 10^–8 x $\dfrac{l}{19.625 \times 10^{-8}}$

l = $\dfrac{10 \times 19.625}{1.6 }$ = 122.7 m

The length of the wire is 122.72 m.

When the diameter is doubled

According to the formula,

R = ρ $\dfrac{l}{A}$ where A = $\dfrac{πd^2}{4}$

Hence, R ∝ $\dfrac{1}{d^2}$

When diameter is double,

Resistance will change by $\dfrac{1}{(2d)^2}$ = $\dfrac{1}{4d^2}$

∴ Resistance will become $\dfrac{1}{4}^{\text{th}}$ of its original value.

Hence, new resistance = $\dfrac{1}{4} \times 10$ = 2.5 Ω

Question 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 –

Plot a graph between V and I and calculate the resistance of that resistor.

Answer

The graph is plotted as below:

The slope of the line indicates the value of resistance.

Let P₁(3.4, 1) and P₂(10.2, 3) be two points on the graph

Slope = $\dfrac{1}{\text{R}} = \dfrac{3-1}{10.2-3.4} = \dfrac{2}{6.8}$

R = $\dfrac{6.8}{2}$ = 3.4 Ω

Therefore, the resistance = 3.4 Ω.

Question 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.

Answer

Given,

V = 12 V

I = 2.5 mA = 2.5 x 10^-3A

R = ?

According to Ohm's Law:

V = IR

$\therefore \text{R} = \dfrac{\text{V}}{\text{I}} \\[1em] \Rightarrow \text{R} = \dfrac{12}{2.5 \times 10^{-3}} \\[1em] \Rightarrow \text{R} = 4.8 \times 10^{3} \Omega$

Hence, resistance of the resistor = 4.8 kΩ

Question 9

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 resistor?

Answer

Given,

V = 9V

Resistors in series = 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω,

R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

According to Ohm's Law:

V = IR

$\therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{9}{13.4} \\[1em] \Rightarrow \text{I} = 0.67 \text{ A}$

Therefore, current flowing = 0.67 A.

Question 10

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

Answer

Given,

I = 5 A

V = 220 V

Let number of resistors be n

Resistance of each resistor = 176 Ω

$\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{176} + \dfrac{1}{176} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{176} \Omega \\[1em] \Rightarrow \text{R}_{eq} = \dfrac{176}{\text{n}} \Omega$

Applying Ohm's Law:

V = IR

220 = 5 x $\dfrac{176}{\text{n}}$

n = $\dfrac{176 \times 5}{220}$ = 4 resistors

Hence, 4 resistors are connected in parallel.

Question 11

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.

Answer

If we connect resistors in series, R = 6 Ω + 6 Ω + 6 Ω = 18 Ω.

If we connect all resistors in parallel,

$\dfrac{1}{\text{R}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

Hence, R = 2Ω

We can obtain the desired value by connecting two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence, R_p = 3Ω

The third resistor in series, then we get equivalent resistance

R_p + 6 = 3 + 6 = 9Ω

∴ To get a resistance of 9 Ω, two 6 Ω resistors should be connected in parallel and the third 6 Ω resistor should be connected in series with the combination as shown below:

Case (ii)

When two resistors are connected in series, their equivalent resistance is

R_s = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is:

$\dfrac{1}{\text{R}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4}$

⇒ R = 4 Ω

∴ To get a resistance of 4 Ω, two 6 Ω resistors should be connected in series and the third 6 Ω resistor should be connected in parallel with the combination as shown below:

Question 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?

Answer

Given,

V = 220 V

P = 10 W

I_Max = 5 A

The resistance of the each bulb:

R = $\dfrac{\text{V}^2}{\text{P}} = \dfrac{220^2}{10} = 4840 Ω$

Assuming n bulbs are connected in parallel

$\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{1}{4840} + \dfrac{1}{4840} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{4840} \Omega \\[1em] \Rightarrow \text{R}_{eq} = \dfrac{4840}{\text{n}} \Omega$

Applying Ohm's Law:

V = IR

220 = 5 x $\dfrac{4840}{\text{n}}$

n = $\dfrac{4840 \times 5}{220}$ = 110 lamps

Hence, 110 lamps can be connected in parallel.

Question 13

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

Answer

Case 1 — When coils are used separately

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{24} \\[1em] \Rightarrow \text{I} = 9.166 \text{ A} \approx 9.2 \text{ A}$

Hence, 9.2 A of current flows through each resistor when they are used separately.

Case 2 — When coils are connected in series

R_s = 24 Ω + 24 Ω = 48 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{48} \\[1em] \Rightarrow \text{I} = 4.58 \text{ A} \approx 4.6 \text{ A}$

Therefore, the current is 4.6 A when the two coils are used in series.

Case 3 — When coils are connected in parallel

$\dfrac{1}{\text{R}_p} = \dfrac{1}{24} + \dfrac{1}{24} = \dfrac{2}{24} = \dfrac{1}{12}$

∴ R_p = 12 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{12} \\[1em] \Rightarrow \text{I} = 18.3 \text{ A}$

Therefore, the current is 18.3 A when the two coils are used in parallel.

Question 14

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 resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series:

R_s = 1 Ω + 2 Ω = 3 Ω.

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{6}{3} \\[1em] \Rightarrow \text{I} = 2 \text{ A}$

Power across 2 Ω resistor = I²R = 2 x 2 x 2 = 8 W

Hence, the power consumed by the 2 Ω resistor is 8 W.

(ii) In parallel circuit, voltage across the resistors remains the same. So, voltage across 2 Ω resistor is 4 V:

Power across 2 Ω resistor = $\dfrac{\text{V}^2}{{\text{R}}} = \dfrac{4 \times 4}{2}$ = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Hence, same power is consumed by 2 Ω resistor in both the circuits.

Question 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?

Answer

As the two lamps are connected in parallel, the voltage across each of them will be the same.

Let I₁ be the current drawn by 100 W bulb:

I₁ = $\dfrac{\text{P}}{\text{V}} = \dfrac{100}{220} = \dfrac{5}{11} \text{ A}$

Let I₂ be the current drawn by 60 W bulb:

I₂ = $\dfrac{\text{P}}{\text{V}} = \dfrac{60}{220} = \dfrac{3}{11} \text{ A}$

∴ Current drawn from line = I₁ + I₂

$= \dfrac{5}{11} + \dfrac{3}{11} \\[1em] = \dfrac{5 + 3}{11} \\[1em] = \dfrac{8}{11} \text{ A} \\[1em] = 0.727272 \text{ A} \approx 0.73 \text{ A}$

Hence, 0.73 A current flows through the line.

Question 16

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

Answer

Given,

Power of TV set = P₁ = 250 W

Time taken by TV set = t₁ = 1 hr = 3600 s

Power of toaster = P₂ = 1200

Time taken by toaster = t₂ = 10 minutes = 600 s

We know, energy consumed (H) = Pt

Energy consumed by TV set:

E₁ = 250 × 3600 = 900000 J = 9 × 10⁵ J

Energy consumed by toaster:

E₂ = 1200 × 600 = 720000 = 7.2 × 10⁵ J

As E₁ > E₂, hence, energy consumed by the TV set is greater than the toaster.

Question 17

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.

Answer

Given,

I = 15 A

R = 8 Ω

The rate at which the heat develops in the heater (P) = I²R

Substituting the values in the equation, we get

P = 15 x 15 x 8 = 1800 watt

Therefore, electric heater produces heat at the rate of 1800 W

Question 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?

Answer

(a) The resistivity and melting point of tungsten is very high hence it doesn't burn readily when heated. Therefore, tungsten is used almost exclusively for filament of electric lamps.

(b) Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices like electric iron, toasters etc.

(c) The series arrangement is not used for domestic circuits due to the following reasons:

1. All the connected appliances cannot be operated independently. If one device is defective, then the entire circuit will not function.
2. The overall voltage gets distributed in a series circuit. As a result, electric appliances may not get the rated power for their operation.
3. The total resistance becomes large, hence, the current is reduced.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when the area of cross section increases the resistance decreases and vice versa.
R = ρ $\dfrac{l}{A}$

(e) Copper and aluminium both are good conductors of electricity and have low resistivity. Hence, they can efficiently carry electrical current with minimal resistance, which reduces energy losses as heat during transmission.