NCERT Exercise Solution
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
Answer: (d) 25
Explanation: The piece of wire having resistance equal to R is cut into five equal parts. Therefore, resistance of each part would be R/5.
When all parts are connected in parallel, the resistance of total resistance can be given as follows:
`(1)/(R') = 5 xx (5/R) = (25)/(R)`
Or, `(R)/(R) = 25`
Question 2: Which of the following terms does not represent electrical power in a circuit?
Answer: (b) IR2
Explanation: We know that Power (P) = VI
After substituting the value of V = IR in this we get
`P = (IR) I = I xx R xx I = I^2R`, Thus `P = I^2R`
Question 3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
- 100 W
- 75 W
- 50 W
- 25 W
Answer: (d) 25 W
Explanation: Potential difference, V = 220 V, Power, P = 100 W
Therefore, power consumption at 100 V =?
To solve this problem, first of all resistance of the bulb is to be calculated.
We know that `P = V^2 ÷ R`
Or, `100 W = (220 V)^2 ÷ R`
Or, `R = 48400 ÷ 100 = 484 Ω`
Now, when the bulb is operated at 110 V, then power can be calculated as follows:
`P = 110^2 ÷ 484 = 12100 ÷ 484 = 25 W`
Thus, bulb will consume power of 25W at 110V
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
Answer: (d) 4 : 1
Explanation: Let the potential difference = V,
Resistance of the wire = R
Resistance when the given wires connected in series = Rs
Resistance when the given wires connected in parallel = Rp
Heat produced when the given wires connected in series = Hs
Heat produced when the given wires connected in parallel = Hp
Thus, resistance Rs when the given two wires connected in series = R + R = 2R
Resistance Rp when the wires are connected in parallel can be calculated as follows:
`1/R_p = 1/R + 1/R = 2/R`
Or, `R_p = R/2`
We know, heat produced `H = I^2R t`
Ratio of heat produced in two conditions:
`H_s : H_p = 2R ÷ R/2 = 4 : 1`
Question 5: How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: Voltmeter is connected into parallel to measure the potential difference between two points in a circuit.
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 of wire = 0.5 mm
Hence, radius `= 0.25 mm = 0.00025 m`
Resistivity, `ρ = 1.6 xx 10^(-8) Ω m`
Resistance (R) = 10 Ω and length = ?
Resistance (R1) when diameter is doubled = ?
When diameter is doubled, radius becomes 0.0005 m
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 slope of the graph will give the value of resistance.
Let us consider two points A and B on the slope.
Draw two lines from B along X-axis and from A along Y-axis, which meets at point C
Now, BC = 10.2 V – 3.4 V = 6.8 V
AC = 3 – 1 = 2 ampere
Thus, resistance `R=3.4Ω`
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, Potential difference, V = 12 V
Current (I) across the resistor `= 2.5mA = 2.5 xx 10 -3 = 0.0025 A`
Resistance, R =?
We know, `R = V/I`
`= 12 V ÷ 0.0025 A = 4800 Ω`
Question 9: A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer: Given, potential difference, V = 9 V
Resistance of resistors which are connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω respectively
Current through resistor having resistance equal to 12Ω =?
Total effective resistance, `R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω`
We know, `I = V/R`
`= 9 V ÷ 13.4 Ω = 0.671 A`
Since, there is no division of electric current, in the circuit if resistors are connected in series, thus, resistance through the resistor having resistance equal to 12 Ω = 0.671 A