# Electricity

## NCERT Exercise Solution

### Part 1

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

- `(1)/(25)`
- `1/5`
- 5
- 25

**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?

- I
^{2}R - IR
^{2} - VI
- V
^{2}/R

**Answer:** (b) IR^{2}

**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

- 1:2
- 2:1
- 1:4
- 4:1

**Answer:** (d) 4 : 1

**Explanation:** Let the potential difference = V,

Resistance of the wire = R

Resistance when the given wires connected in series = R_{s}

Resistance when the given wires connected in parallel = R_{p}

Heat produced when the given wires connected in series = H_{s}

Heat produced when the given wires connected in parallel = H_{p}

Thus, resistance R_{s} when the given two wires connected in series = R + R = 2R

Resistance R_{p} 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 (R_{1}) when diameter is doubled = ?

We know,

`R=ρl/A`

So, `l=(RA)/(ρ)=(Rπr^2)/(ρ)`

Or, `l=(10Ωxx3.14xx(0.00025m)^2)/(1.6xx10^(-8)Ωm)`

`=(10Ωxx3.14xx0.00025mxx0.00025m)/(1.6xx10^(-8)Ωm)`

`=(196.25)/(1.6)m=122.656 m`

When diameter is doubled, radius becomes 0.0005 m

So, `R_1=ρl/A=ρ(l)/(πr^2)`

`=(1.6xx10^(-8)xx122.7)/(3.14xx0.00000025xx10^8)Ω`

`=(196.32)/(78.5)Ω=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

I(ampere) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
---|---|---|---|---|---|

V(volt) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

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

Slope `=1/R=(AB)/(BC)=(2)/(6.8)=(1)/(3.4)`

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