Q: 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.

Plot of I vs V is shown in the figure.Resistance=Slope1From the graph, Slope=3.4−1.61−0.5=0.278 Ω−1 Slope=3.4−1.61−0.5=0.278 Ω−1

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

For resistors in series, Req=R1+R2+R3+R4+R5 =0.2+0.3+0.4+0.5+12 =13.4 Ω By ohm's Law:V=IReq9=13.4II=0.67 AWhen resistors are connected in series, current is same in all the resistors. Hence, current in 12 Ω resistor =0.67 A.

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 R′R​ is -

Accepted Answer

Let the length of resistance R be LLet the resistance of each part after cutting be Rs​Since resistance is proportional to the length of the resistor,Rs​=R/5.....................(i)For resistors in parallel, R′1​=R1​1​+R2​1​+R3​1​+R4​1​+R5​1​⟹R′=Rs​/5................(ii)From (i) and (ii),R′=R/25R/R′=25

Question 2

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

Accepted Answer

IR2 does not represent electrical power in a circuit.

Question 3

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

Accepted Answer

P=V2/RAt 220 V,100=2202/R..........(i) At 110 V,P=1102/R...........(ii)Dividing (i) from (ii),P/100=1102/2202P=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 in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :

Accepted Answer

Since the two wires have same material, length and diameter, they have the same resistance. Let the resistance of each wire be R.Let applied potential difference be V.Rseries​=R1​+R2​             =R+R=2RPseries​=V2/Rseries​             =V2/(2R)..........(i)Rparallel​1​=R1​1​+R2​1​Rparallel​=R/2Pparallel​=V2/Rparallel​              =2V2/R............(ii)Dividing (ii) from (i),Pparallel​Pseries​​=1/4

Question 5

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

Accepted Answer

A voltmeter is connected in parallel to the two points to measure the potential difference between the two points.

Question 6

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

Accepted Answer

Given, diameter, d=0.5 mm            resistivity, ρ=1.6×10−8 Ωm            Resistance, R=10 ΩLet the length of wire be l.     A=πd2/4     R=ρl/A              =πd2/4ρl​⟹l=4ρRπd2​             l=4×1.6×10−810×3.14×(0.5×10−3)2​             l=122.7 m         R∝1/d2If the diameter is doubled, resistance will be one-fourth.Hence, new resistance =2.5Ω

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.

Accepted Answer

Plot of I vs V is shown in the figure.Resistance=Slope1​From the graph, Slope=3.4−1.61−0.5​=0.278 Ω−1 Slope=3.4−1.61−0.5​=0.278 Ω−1

Question 8

When a 12V 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.

Accepted Answer

Given: V=12 V            I=2.5 mALet the resistance be RBy Ohm's Law, V=IR12=2.5×10−3RR=4.8×103 Ω

Question 9

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

Accepted Answer

For resistors in series, Req​=R1​+R2​+R3​+R4​+R5​       =0.2+0.3+0.4+0.5+12       =13.4 Ω  By ohm's Law:V=IReq​9=13.4II=0.67 AWhen resistors are connected in series, current is same in all the resistors. Hence, current in 12 Ω resistor =0.67 A.

Question 10

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

Accepted Answer

Given: Individual resistance, R=176 Ω                                                I=5 A                                                V=220 V   Let number of resistors be n.Then, R1​=R2​=......=Rn​=R=176 Ω............(i)For resistors in parallel, Req​1​=R1​1​+R2​1​+.....Rn​1​..........(ii)From (i) and (ii),Req​=176/n...........(iii)By ohm's Law,V=IReq​220=5Req​...............(iv)Substituting (iii) in (iv),220=5×176/nn=4Hence, four resistors are required.

Question 11

Show how you would connect 3 resistors , each of resistance 6 ohm, so that the combination has a resistance of (i) 9 ohm (ii) 4 ohm

Accepted Answer

We are given 3 resistors each of resistance 6 Ω. Equivalent resistance of resistors in series is more than the individual resistance of each resistor while equivalent resistance of resistance in parallel is less than the individual resistance of each resistor. (i)To get an equivalent resistance of 9 Ω, atleast one six ohm resistor should be connected in series with the other resistors. Since, Req​=R1​+R2​9=6+R2​R2​=3 Ω⟹R2​<6 ΩConnecting the remaining two resistors in parallel will yield an equivalent resistance of 3 Ω.Hence, the first figure shows the configuration of resistors to get equivalent resistor of 9 Ω. (ii)To get an equivalent resistance of 9 Ω, no six ohm resistor should be connected in series with the other resistors. Let one of the branches have only one 6 Ω resistor.Since, Req​1​=R1​1​+R2​1​91​=61​+R2​1​R2​=12 ΩConnecting the remaining two resistors in series will yield an equivalent resistance of 12 Ω.Hence, the second figure shows the configuration of resistors to get equivalent resistor of 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?

Accepted Answer

Given, V=220 V            Power rating of each bulb, P=10 W            Total current, It​=5 ALet number of lamps in parallel be n.Let individual resistance of each lamp be R.From Joule's law of heating,P=V2/R10=2202/RR=4840.............(i)R1​=R2​=..............=Rn​=R..............(ii)For resistances in parallel, Req​1​=R1​1​+R2​1​+......+Rn​1​........(iii)From (ii) and (iii),Req​=R/n..........(iv)Substituting (i) in (iv),Req​=4840/n...............(v)From Ohm's Law,V=It​Req​............(vi)Substituting (v) in (vi),220=5×4840/nn=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?

Accepted Answer

Case 1: Coils used separately.       V=IR       220=24I          I=9.167 ACase 2: Coils used in series       Req​=R1​+R2​               =24+24=48 Ω       V=IReq​       220=48I       I=4.583 ACase 3: Coils used in parallel       Req​1​=R1​1​+R2​1​         Req​=12 Ω         V=IReq​       220=12I       I=18.334 A

Question 14

Calculate 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.

Accepted Answer

(i)For resistors in series,Req​=R1​+R2​Req​=1+2=3 ΩFrom Joule's Law of heating,P=V2/Req​     =62/3     =12 W(ii)For resistors in parallel,Req​1​=R1​1​+R2​1​Req​=12/7 ΩFrom Joule's Law of heating, P=V2/Req​    =12/742​    =9.33 WAs calculated, power  used in case (i) is more than power used in case (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 voltage is 220 V?

Accepted Answer

From Joule's Law of heating,P=V2/RFor first lamp, R1​=2202/100For second lamp, R2​=2202/60When they are connected in parallel,Req​1​=R1​1​+R2​1​=2202/1001​+2202/601​⟹Req​=2202/160From Ohm's Law,V=IReq​220=1602202​II=0.73 A

Question 16

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

Accepted Answer

E=PtFor TV,E1​=250×1=250 WhrFor toaster,E2​=1200×6010​=200 WhrHence, TV uses more energy.

Question 17

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

Accepted Answer

Rate of generating heat is equal to the power used by the heater.By Joule's Law of heating, P=I2R    =152×8    =1800 WHence, rate of generation of heat is 1800W.Total amount of heat generated, E=Pt E=1800×2=3.6 kWh

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 aluium wires usually employed for electricity transmission?

Accepted Answer

(a) Electric lamps heat up with time as they are used. Since, tungsten has very high melting point, it does not melt even at very high temperatures. Hence, the lamp in not destroyed.(b) Electrical heating devices are required to produce a lot of heat and maintain high temperatures. Since, heat generated is proportional to resistivity of device, materials of higher resistivities are preferred in the construction of heating devices. Alloys have more resistance than metals and hence are preferred for the same.(c) When resistors are connected in series, potential difference across each resistor is smaller than the applied voltage. Hence, the power delivered to each device is reduced and hence they don't work at full efficiency.(d)R=ρl/AHence, resistance is inversely proportional to the cross section of the resistance of the wire.(e) Cables used for electrical transmission should use up as small power as possible so that most of the power is delivered to the electrical equipment. Power consumed in the wire is proportional to resistivity. Copper and aluium have very low resistivity and hence well suited for the purpose of being used as wires.

Question 19

An electric heater is rated at 2 kW. Electrical energy costs Rs 4 per kWh. What is the cost of using the heater for 3 hours?

Accepted Answer

Electrical energy consumed in three hours =P×t=2 kW×3 hour =6kWhUnit cost = Rs. 4 per kWhTherefore, the cost of energy used for three hours =4×6= Rs. 24

Question 20

When a 4Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second is:

Accepted Answer

The number of coulombs passing through the resistor is the current passing through it.Current = Voltage/ResistanceI=V/RI=12/4I=3 AThus, when a 4Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second will be 3 .

Filters

Filters

class

subject

topic

concept

difficulty

Solutions for Electricity Important Questions

Found the solution, but did not understand the concept?

Ask my first question