The inherent property of a conductor because of which it resists the flow of electric current is called resistivity. Resistivity for a particular material is unique. Resistance varies directly as length of the conductor. Current varies inversely as resistance. So, when length of the wire is doubled, its resistance becomes double. When resistance becomes double, current becomes half. This explains why the reading of ammeter decreases to half when the length of the wire is doubled.

The commercial unit of electrical energy is kilowatt hour.
 1 kWh = 11000 W h
 1000 W h= (1000 Joule/second) ×hour 
 
 = 1000 x J / S x 60 x 60 s 
 = 3600 x 1000 J 
 = 3.6 x 10⁶ J

Total resistance of circuit can be calculated as follows: 

R= V / I 

= 10 / 1

= 10  Ω

Since lamp and conductor are in series so resistance of lamp =10 Ω − 5Ω = 5Ω .The new resistance in parallel to earlier combination has same value, i.e. 10Ω as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 5Ω conductor. Now, resistance remains the same but current has become half. Using Ohm formula, potential difference across the lamp can be calculated as follows: V = IR = 0.5A x 5Ω = 2.5V

In parallel arrangement, total resistance of the circuit is less than arithmetic sum of resistors in the wiring. This helps in reducing the load on the wiring. In parallel arrangement, each appliance can have its independent switch. Fault in one section is not going to affect any other section. These advantages make parallel arrangement ideal of domestic wiring

B1, B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
Ans. In parallel circuit, potential difference does not get divided. Hence, glow of other bulbs will not be affected when bulb B1 gets fused.

(ii) What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
Ans. Ammeter A shows a reading of 3A. This means each of the A1, A2, and A3 show 1A reading.
When the bulb B2 gets fused, no current flows through this bulb. So, all the current is equally divided between remaining two bulbs. So, ammeters A1 and A2 will show 3/2 = 1.5 A current each. Ammeter A3 will show zero current.

(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
Ans. For finding power, we need to first calculate the resistance in the circuit

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
Ans. Voltage gets divided in series combination, So bulbs in series combination will glow with less brightness. Voltage does not get divided in parallel combination, so bulbs in parallel combination will glow with more brightness.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
Ans. In case of series combination, fault in even a single component will break the circuit.
So, when one of the bulbs gets fused; another bulb would continue to glow.

Ohm’s Law: At constant temperature, potential difference across a conductor is directly proportional to electric current passing through it. If V is the potential difference and I is electric current, then as per Ohm’s law;

VαI

or V/ I = R

Where R is the constant of proportionality and it is called resistance.

Verifying Ohm’s Law:

• Take a nichrome wire, some electric cells, key, voltmeter and ammeter.
• Connect the component to make a circuit; as shown in given figure.
• Start with 1 cell in the circuit. Switch on the key and take the reading of ammeter.
• Increase the number of cells to 2. Switch on the key and take the reading of
ammeter.
• Increase the number of cells in similar increments and take the reading of
ammeter in each case.
Tabulate your data and plot a V-I graph.

The graph shows that potential difference varies directly as electric current. Ohm’s law holds good under normal conditions. But, it may not hold good under exceptional conditions.

The inherent property of a conductor because of which it resists the flow of electric
current is called resistivity. Resistivity for a particular material is unique.
The SI unit of resistivity is Ωm (Ohm metre).

Experiment to study the factors on which resistance of conducting wire depends:
• Take an ammeter, electric cell, plug key, nichrome wire and wires of different materials.
• Make the circuit as shown in figure.
• Start the experiment with nichrome wire. Attach it in the circuit and take ammeter reading.
• Change the length of nichrome wire and take ammeter reading.
• Change the thickness of nichrome wire and take ammeter reading.
• After above steps, use copper wire for the experiment. Attach a copper wire in the circuit and take ammeter reading.
• Change the length of copper wire and take ammeter reading.
• Change the thickness of copper wire and take ammeter reading.
• Repeat above steps with wires of different materials. 

 

Observations:
• It is seen that resistance depends on material of conductor.
• Resistance depends on length of conductor.
• Resistance depends on area of cross-section.