## Class Ten Science Chapter 12 NCERT Solution 2

#### **Chapter 12: Electricity**

**Q.12. Draw a schematic diagram of a circuit consisting of a
battery of three cells of 2 V each, a 5 **Ω **resistor,
an 8 **Ω **resistor, and a 12 **Ω **resistor, and a plug key, all
connected in series. **

**Ans. **Three cells of potential 2 V, each connected in series, therefore, the potential difference of the battery will be 2 V + 2 V + 2 V = 6 V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.

**Q.13. Redraw the circuit of question 12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 **Ω **resistor. What would be the readings in the ammeter and the voltmeter? **

**Ans. **An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.

The resistance is connected in a series.

Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,

V = IR,

where,

Potential difference, V = 6 V

Current flowing through the circuit (resistors) = I

Resistance of the circuit,

R = 5 + 8 + 12 = 25 Ω

I = V/R = 6/ 25 = 0.24 A

Potential difference across 12 Ω restore = V1

Current flowing through the 12 Ω resistor, I = 0.24 A

Therefore, using Ohm’s law, we obtain

V1 = IR = 0.24 × 12 = 2.88 V

Therefore, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter will be 2.88 V.

**Q.14. Judge the equivalent resistance when the following are
connected in parallel : **

**(a) 1 **Ω **and 106 **Ω**, (b)
1 **Ω **and 103 **Ω **and 106 **Ω**. **

**Ans. **(a) When 1 Ω and 106 Ω are
connected in parallel :

Let R be the equivalent resistance.

Equivalent resistance in a parallel combination of resistors is always less than the least resistance of any resistors in the circuit.

Hence, in both the cases i.e., (a) and (b) the equivalent resistance is less than 1Ω.

**Q.15. An electric lamp of 100 **Ω**,
a toaster of resistance 50 **Ω **and a water filter of resistance 500 **Ω **are
connected in parallel to a 220 V source. What is the resistance of an electric
iron to the same source that takes as much current as all three appliances, and
what is the current through it? **

**Ans. The resistance** of electric lamp, R1 = 100 Ω

The resistance of toaster, R2 = 50 Ω

The resistance of water filter, R3 = 500 Ω

The potential difference of the source,

V = 220 V

These appliances are connected in parallel.

Let R be the equivalent resistance of the circuit.

7.04 A of current is drawn by all the three given appliances.

Therefore, currents drawn by an electric iron are connected to the same source of potential

220 V = 7.04 A

Let R′ be the resistance of the electric iron. According to Ohm’s law,

V = IR′

R′ = V/I = 220 /7.04 .

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.