[**NCERT In Text**]

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.

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 = 6V. 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.

[**NCERT In Text**]

Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

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 resistances are connected in 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 = 25Ω

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

Potential difference across 12 Ω resistor = V1

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

Therefore, using Ohm’s law, we obtain

V1 = IR = 0.24 x 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.

[**NCERT In Text**]

Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.

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

Let R be the equivalent resistance.

Therefore, equivalent resistance ≈ 1 Ω

(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:

Let R be the equivalent resistance.

Therefore, equivalent resistance = 0.999 Ω

Let R be the equivalent resistance.

Therefore, equivalent resistance ≈ 1 Ω

(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:

Let R be the equivalent resistance.

Therefore, equivalent resistance = 0.999 Ω

[**NCERT In Text**]

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 Vsource. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Resistance of electric lamp, R1 = 100 Ω

Resistance of toaster, R2 = 50 Ω

Resistance of water filter, R3 = 500 Ω

Potential difference of the source, V = 220 V

These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.

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

Resistance of toaster, R2 = 50 Ω

Resistance of water filter, R3 = 500 Ω

Potential difference of the source, V = 220 V

These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.

According to Ohm's law,

V = IR

Where,

Current flowing through the circuit = I

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

Therefore, current drawn by an electric iron 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,

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