Electricity

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–

(a) 1/25

(b) 1/5

(c) 5

(d) 25

Answer: (d) 25

Explanation:

Resistance of whole wire = R

Therefore, resistance of each part of the wire = R/5

Since all the pieces of wires are connected in parallel, and the equivalent resistance of this combination = R

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

(a) I²R

(b) IR²

(c) VI

(d) V²/R

Answer: (b) IR²

Explanation:

Different forms of expression of power

We know that,

P = VI --------(1)

Thus, electric power can be defined by above three forms of expressions, and not by IR²

Thus, (b) IR² is correct answer

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 W

(b) 75 W

(c) 50 W

(d) 25 W

Answer: (d) 25 W

Explanation:

Given,

V = 220V, P = 100W

At 110V, Power consumption = ?

We know that,

Now, Power consumption at 110V,

Thus, power consumption at 110V = 25W

And option (d) 25 W is correct answer.

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) 1 : 4

Solution:

Let, potential difference = V,

Let, resistance of the wire = R

Let resistance of wire in series = R₁

Let resistance of wire in parallel = R₂

Let heat produced in series = H₁

Let heat produced in parallel = H₂

Total effective resistance of wires when connected in series, R₁ = R + R = 2R

Total effective resistance of wires when connected in parallel,

By substituting the value of R₂ and R₁

Therefore, option (c) 1 : 4 is the correct answer.

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 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,

Resistance, R=10 Ω

Resistivity (ρ)= 1.6 × 10^–8 Ω m

Diameter = 0.5mm

Therefore, length =?

And, Resistance =? (When the diameter is doubled.)

Here, diameter of wire = 0.5 mm

Therefore, radius = 0.5mm/2 = 0.25mm = 0.25/1000 m = 0.00025m

Now,

Area of cross section (A) = π r²

Or, A = 3.14 x 0.00025 m x 0.00025 m

When diameter of wire become double, i.e. equal to 0.5 mm x 2 = 1 mm

Therefore, radius, r = 1mm/2 = 0.5mm = 0.5/1000 m = 0.0005 m

Therefore, Area A = πr²

⇒ A = 3.14 x 0.0005 m x 0.0005 m

⇒ A = 3.14 x 0.00000025 m²

Thus, length of the given wire = 122.65 meter

Resistance of the wire when diameter becomes double = 2.5Ω