### Intext Questions Solved

#### Page 200

**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 :** When one coulomb of charge flows through a conductor in one second then current is one ampere.

**Question 3 : Calculate the number of electrons constituting 1 coulomb of charge.**

**Answer :** Charge on one electron, e = 1.6 x 10^{-19} coulomb, Q = 1 coulomb.

#### Page 202

**Question 1 : Name a device that helps to maintain a potential difference across a conductor.**

**Answer :** A cell or a battery.

**Question 2 : What is meant by saying that the potential difference between points is 1 V?**

**Answer :** If one J of work is done in moving one coulomb (1 C) charge from one point to another point then potential difference between the points is one volt.

**Question 3 : How much energy is given to each coulomb of charge passing through a 6 V battery?**

**Answer :** We know that

V = W/Q or W=VQ

W = 6 x 1 = 6 joule

#### Page 209

**Question 1 : On what factors does resistance of a conductor depend?**

**Answer :** The resistance (R) of conductor depend upon the following factors.

i) Its length, l R ∝ l.

ii) Its area of cross-section, A (R ∝ 1/A)

iii) Its resistivity of material , ρ

iv) Its temperature

**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 :** Current will flow more easily through a thick wire as resistance of a thick wire is less than that of a thin wire (R ∝ 1/A).

**Question 3 : Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its formal value. What change will occur in the current through it?**

**Answer :** We know that I = V/R

If V is halved and R remains constant then current will also reduce to half to I/2, i.e.

**Question 4 : Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal?**

**Answer :** There are two main reasons for this:

i) Resistivity of an alloy is generally higher than that of its constituents metals, and

ii) Alloys do not oxidise (burn) readily at high temperatures.

**Question 5 : Answer the following questions:**

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

**b) Which material is the best conductor?**

**Answer :** a) Iron is better conductor as its resistivity is less than that of mercury.

b) Silver.

#### Page 213

**Question 1 : Draw a schematic diagram of a circuit consisting of a battery of three cells, 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 :**

As 5 Ω , 8 Ω and 12 Ω are in series so the net resistance

R = 5 + 8+ 12

R = 25 Ω

Now, I = V/R

I = 6/25

I = 0.24 A

Page 216

**Question 1 : Judge the equivalent resistance when the following are connected in parallel: a) 1 Ω and 10 ^{6} Ω (b) 1 Ω and 10^{3} Ω and 10^{6} Ω.**

**Answer :** a) Equivalent resistance will be less than 1 Ω as in parallel combination the net resistance is less than the least resistance.

b) Equivalent resistance will be less than 1 Ω due to same reason as above.

**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 200 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 : Resistance of electric lamp, R1 = 100Ω

Resistance of toaster , R2 = 50Ω

Resistance of water filter, R3 = 500 Ω

Supply voltage, V = 220 V

As these appliances are connected in parallel, the net resistance R is given as

**Question 3 : What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?**

**Answer :** Advantages are as under:

i) each appliance gets the same voltage as the source voltage and current divides according to the requirement of the appliance.

ii) even if one appliance fails other appliances will be working.

iii) In parallel combination the net resistance is decreased and each appliance having difference resistance can operate properly.

**Question 4 : How can three resistors of resistances 2 Ω, 3 Ω, and 6Ω be connected to give a total resistance of (a) 4Ω and (b) 1 Ω?**

**Answer :** (a) To get a net resistance of 4 Ω connect 3 Ω and 6 Ω in parallel and 2 Ω in series with this combination.

**Question 5 : What is (a) the highest, (b) the lowest total, resistance that can be secured by combination of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?**

**Answer :** a) The highest resistance is obtained when resistors are connected in series,

#### Page 218

**Question 1 : Why does the cord of an electric heater not glow while the heating element does?**

**Answer :** The cord of heater is made of copper or aluminium which has low resistance (R) so the heat produced (I^{2}Rt) is very small. But the element of heating coil is made of alloy whose resistance is very high due to which the heat produced (I^{2}Rt) is very high and heater element becomes hot and glows.

**Question 2 : Compute the heat generated while transferring 96000 C of charge in one hour through a potential difference of 50 V.**

**Answer :** We are given : Q = 96000 C, t = 1 h, V = 50 V

Heat produced, H = V/t = VxQ

H = 50 x 96500

= 4.8 x 10^{6 }J.

**Question 3 : An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.**

**Answer :** We have, R = 20 Ω, I = 5A, t = 30 s

Now, H = I^{2}Rt

H = (5)^{2} x 20 x 30 = 1.5 x 10^{4} J.

#### Page 220

**Question 1 : What determines the rate at which energy is delivered by a current?**

**Answer :** Electric power.

**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 hour.**

**Answer :** We are given,

I = 5 A

t = 2 h = 2 x 60 x 60 = 7200 s

V = 220 V

Now, Power, P = VI = 220 x 5 = 1100 watt.

Energy = V/t = P x t = 1100 x 7200 = 7.92 x 10^{6} J.

### Text Book Questions Solved

**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 ration R:R’ is**

a) 1/25

b) 1/5

c) 5

d) 25

**Answer :** (d) as each cut part has resistance R/5 and their parallel combination has resultant R’ which is given as:

**Question 2 : Which of the following terms does not represent electrical power in a circuit?**

a) I^{2}R

b) IR^{2}

c) VI

d) V^{2}/R

**Answer :** (b) IR^{2}

**Question 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 Ω

b) 75 Ω

c) 50 Ω

d) 25 Ω

**Answer :** (d)

Resistance of bulb,

R = V^{2}/P

= (220)^{2}/100

= 484 Ω

When connected to 110 V, the power consumed = (110)^{2}/848 = 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**

a) 1:2

b) 2:1

c) 1:4

d) 4:1

**Answer :** (c)

Let the resistance of each wire be R then:

In series combination:

**Question 5 : How is a voltmeter connected in the circuit to measure the potential difference between two points?**

**Answer :** A voltmeter is connected in parallel to the component across which potential difference is to be measured.

**Question 6 : A copper wire has diameter 0.5 mm and resistivity of 1.6 x 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

= 0.5/10 cm

= 0.05 cm

So, radius , r = 0.05/2 = 0.025 cm

Resistivity, ρ = 1.6 x 10^{-6} Ω m

Resistance, R = 10 ohm

Length, l = ?

R = V/l

Using R= ρ (l/A) or l = RA/ρ

(where, A = πr^{2})

So, if diameter is doubled then resistance will become four times.

**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 :** To pot graph between potential difference (V) and current (I), we take a graph paper and values of V are plotted on vertical axis and that of current are taken on horizontal axis taking scale as:

1 cm = 1 ampere (x-axis)

1 cm = 2 volt (y-axis)

R = V/I

R = 6.7/2 (From given Table)

R = 3.35 W

**Question 8 : When a 12 V battery is connecting 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 volt

I = 2.5 mA

Using, R = V/I

R = 12 / (2.5 x 10^{-3})

= 4800 Ω

**Question 9 : A battery 9 V is connected in series with resistors of 20 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 = 9 volt

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

R = 13.4 Ω

Current, I = V/R

I = 9/13.4

I = 0.67 A

As current in series connection is the same through each resistor, therefore , current through 12 Ω will be 0.67 A.

**Question 10 : How many 176 Ω resistors (in parallel) are required to carry a 5 A on a 220 V line?**

**Answer :** Let the number of 176 Ω resistors be n. As these are connected in parallel the resultant resistance will be:

Number of 176 Ω resistors required = 4.

**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 :** (i) Two 6 Ω resistors in parallel would give a resultant of 3 Ω and this combination connected in series with 6 Ω would give 9 Ω.

ii) Three 6 Ω resistors in parallel would give a resultant of 2 Ω and such two combination in series would give 4 Ω.

**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 : P = 10 W, V = 220 volt

Current drawn by each bub = I

I = P/V

I = 10/220 = 1/22 A.

Suppose the number of bulbs are n, as these are connected in parallel, electric current in each would add up hence

n x 1/22 = 5

⇒ n = 110

Number of lamps will be 110.

**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 three cases?**

**Answer :**

Given: V = 220 volt, R_{A} = 24 Ω, R_{B} = 24Ω

i) When used separately, current drawn by each will be the same,

I = V/R = 220/24 = 9.16 A

ii) When connected in series:

Total resistance, R_{S} = R_{A} + R_{B}

R_{S} = 24 + 24 = 48 W

I = V/R_{S} = 220/48 = 4.58 A

iii) When connected 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)

(ii)

**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 voltages is 220 V?**

**Answer :**

**Question 16 : Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?**

**Answer :** Energy used by TV set = P x t

= 250 W x 1 hr

= 0.25 kW x 1 hr. = 0.25 kWh

Energy used by toaster = P x t = 1200 W x 10 min.

= 1.2 kW x 1/6 hr = 0.2 kWh

Energy uses by TV set is more than the energy used by toaster.

**Question 17 : An electric heater of resistance 8 Ω draws 15 A from the service mains in 2 hours. Calculate the rate at which heat is developed in the heater:**

**Answer :** The rate at which heat is developed is power.

P = I^{2}R

P = (15)^{2} x 8

P = 1800 watt.

**Question 18 : Explain the following:**

**a) Why is 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 are of cross-section?**

**e) Why are copper and aluminium wires usually employed for electricity transmission?**

**Answer :** (a) As it has high melting point and it retains heat as much as heat generated and gets very hot and emit light.

(b) These have high resistivity and do not burn (oxidise) readily at higher temperatures.

(c) In series circuit is the same due to which it will be impractical to connect a bulb and an electric heater in series. Also if one component is switched off then other component in series will not work.

(d) R = ρ l/A resistance is inversely proportional to area of cross section .

(e) These have very low resistivity.