Q1 
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 (b) 1/5 (c) 5 (d) 25 
Ans: 
(d) Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts.
Therefore, resistance of each part =
All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as
Therefore, the ratio is 25. 

Q2 
Which of the following terms does not represent electrical power in a circuit?
(a) I2R (b) IR2 (c) VI (d) V2/R 
Ans: 
(b) Electrical power is given by the expression, P = VI … (i)
According to Ohm’s law, V = IR … (ii)
Where,
V = Potential difference
I = Current
R = Resistance
P = VI
From equation (i), it can be written
P = (IR) × I
P = I^{2}R
From equation (ii), it can be written
Power P cannot be expressed as IR^{2}. 

Q3 
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 
Ans: 
(d) Energy consumed by an appliance is given by the expression,
Where,
Power rating, P = 100 W
Voltage, V = 220 V
The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as
Therefore, the power consumed will be 25 W. 

Q4 
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 
Ans: 
(c) The Joule heating is given by, H = i^{2}Rt
Let, R be the resistance of the two wires.
The equivalent resistance of the series connection is R_{S} = R + R = 2R
If V is the applied potential difference, then it is the voltage across the equivalent resistance.
The heat dissipated in time t is,
The equivalent resistance of the parallel connection is
V is the applied potential difference across this R_{P}.
The heat dissipated in time t is,
So, the ratio of heat produced is,
Note: H also H i^{2} and H t. In this question, t is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i^{2}Rt to find the ratio. 

Q5 
How is a voltmeter connected in the circuit to measure the potential difference between two points? 
Ans: 
To measure the potential difference between two points, a voltmeter should be connected in parallel to the points. 

Q6 
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? 
Ans: 
Resistance (R) of a copper wire of length l and crosssection A is given by the expression,
Where,
Resistivity of copper, p = 1.6 x 10^{8} Ω m
Area of crosssection of the wire,
Diameter = 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
Hence, length of the wire,
If the diameter of the wire is doubled, new diameter = 2x0.5 = 1 mm = 0.001 m
Therefore, resistance R^{'}
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω 

Q7 
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 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. 
Ans: 
The plot between voltage and current is called IV characteristic. The voltage is plotted on xaxis and current is plotted on yaxis. 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 10.2 13.2
I (amperes) 0.5 1.0 2.0 3.0 4.0
The IV characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance (R) as,
Therefore, the resistance of the resistor is 3.4 Ω. 

Q8 
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. 
Ans: 
Resistance (R) of a resistor is given by Ohm’s law as,
Where,
Potential difference, V = 12 V
Current in the circuit, I = 2.5 mA = 2.5x10^{3} A
Therefore, the resistance of the resistor is 4.8 kΩ. 

Q9 
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? 
Ans: 
There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 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
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A. 

Q10 
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? 
Ans: 
For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as
Where,
Supply voltage, V = 220 V
Current, I = 5 A
Equivalent resistance of the combination = R, given as
From Ohm’s law,
Therefore, four resistors of 176 Ω are required to draw the given amount of current. 

Q11 
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. 
Ans: 
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be
which is also not desired. Hence, we should either connect the two resistors in series or parallel.
(i) 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.
(ii) Two resistors in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
Therefore, the total resistance is 4 Ω. 

Q12 
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? 
Ans: 
Resistance R_{1} of the bulb is given by the expression,
Where,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P_{1} = 10 W
According to Ohm’s law,
V = I R
Where,
R is the total resistance of the circuit for x number of electric bulbs
Resistance of each electric bulb, R_{1} = 4840 Ω
Therefore, 110 electric bulbs are connected in parallel. 

Q13 
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? 
Ans: 
Supply voltage, V = 220 V
Resistance of one coil, R = 24 Ω
(i) Coils are used separately
According to Ohm’s law,
V = I_{1}R_{1}
Where,
I_{1 }is the current flowing through the coil
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law,
V = I_{2}R_{2}
Where,
I_{2} is the current flowing through the series circuit
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_{3}R_{3}
Where,
I_{3 }is the current flowing through the circuit
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel. 

Q14 
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. 
Ans: 
(i) Potential difference, V = 6 V
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
Where,
I is the current through the circuit
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)^{2}R = (2)^{2} x 2 = 8 W
(ii) Potential difference, V = 4 V
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 resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
Therefore, the power used by 2 Ω resistor is 8 W. 

Q15 
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? 
Ans: 
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,
Similarly, current drawn by the bulb of rating 60 W is given by,


Q16 
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? 
Ans: 
Energy consumed by an electrical appliance is given by the expression,
H = Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 × 3600 = 9 × 10^{5} J
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. 

Q17 
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. 
Ans: 
Rate of heat produced by a device is given by the expression for power as
P = I^{2}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. 

Q18 
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 breadtoasters 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 crosssection?
(e) Why are copper and aluminium wires usually employed for electricity transmission? 
Ans: 
(a) The melting point and resistivity of tungsten are very high. It does not burn readily at a high temperature. The electric lamps glow at very high temperatures. Hence, tungsten is mainly used as heating element of electric bulbs.
(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. It produces large amount of heat.
(c) There is voltage division in series circuits. Each component of a series circuit receives a small voltage for a large supply voltage. As a result, the amount of current decreases and the device becomes hot. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of crosssection (A), i.e.,
(e) Copper and aluminium wires have low resistivity. They are good conductors of electricity. Hence, they are usually employed for electricity transmission. 
