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 - (a) 1/25
ANSWER: - 25
2) Which of the following terms does not represent electrical power in a circuit?
(b) IR 2
(d) V2 /R
ANSWER: - (b) IR 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
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 -
5) How is a voltmeter connected in the circuit to measure the potential difference between two points?
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 × 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:-Area of cross-section of the wire, A =π (d/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that
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
V (volts) 1.6 3.4 6.7 10.
Plot a graph between V and I and calculate the resistance of that resistor.
ANSWER:-The plot between voltage and current is called IV 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 1
I(amperes ) 0.5 1.0 2.0 3
The slope of the line gives the value of resistance ( R ) as,
Slope = 1/R = BC/AC = 2/6.8
R = 6.8/2 = 3.4 Ω
Therefore, the resistance of the resistor is 3.4 Ω.
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.
Therefore, the resistance of the resistor is 4.8 kΩ
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:-V = IR
I= V /R
Where, R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4Ω, 0.5 Ω and 12 Ω. 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 Ω
Potential difference, V = 9 V
I= 9/13.4 = 0.671 A
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
10) How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
ANSWER:-V = IR
R = V /I
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R, given as
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:-Two resistors in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
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:-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=10watts
Because R =V 2/P
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 resistances, which may be use separately, in series, or in parallel. What are the currents in the three? Cases?
ANSWER :-(i) Coils are used separately According to Ohm's law,
V = I 1R 1
I1 is the current flowing through the coil I1 = 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 Ω +24 Ω = 48 Ω
According to Ohm's law, V =I2 R2
I2 is the current flowing through the series circuit I2 = 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
According to Ohm's law,
V = I 3R 3
I3 is the current flowing through the circuit I3 = 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 Ω resistors in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistor, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
ANSWER :-(i) Potential difference, V = 6
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
I is the current through the circuit
I= 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 Ω resistor is 2 A. Power is given by the expression,
P = ( I ) 2R = (2) 2 x 2 = 8 W
(ii) Potential difference, V = 4
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistors will be 4 V.
Power consumed by 2 Ω resistor is given by P = V 2/R = 42 /2 = 8 W
Therefore, the power used by 2Ω 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?
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?
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250×3600 = 9 ×105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
Energy consumed by a toaster of power 1200 W in 10 minutes
= 1200 ×600
= 7.2×10 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 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
ANSWER:-P =I2 R Where, Resistance of the electric heater, R = 8 Ω Current drawn, I = 15 A P = (15) 2 x 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 aluminum wires usually employed for electricity transmission?
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 aluminum are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.