Class 10 Physics

# Electricity

## Electricity: NCERT Solution

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`

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`

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`

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

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`

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.

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`

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`

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`

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

Electric Current

Potential Difference

Ohm's Law

Resistance:1

Resistance: 2

Heating Effect

Applications

Summary

InText Solution -1

InText Solution -2

NCERT Exercise Solution

NCERT Exercise Solution-2

Quiz