Question

In the given circuit, A, B, C and D are four lamps connected with a battery of 60V.

Analyse the circuit to answer the following questions

(i) What kind of combination are the lamps arranged in (series or parallel)?

(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?

(iii) Explain with proper calculations which lamp glows the brightest?

(iv) Find out the total resistance of the circuit.

Answer 1

(i) The lamps are in parallel.

(ii) Advantages:

1. In parallel connection, if one lamp is faulty, it will not affect the working of the other lamps.

2. They will also be using the full potential of the battery as they are connected in parallel.

(iii) In this case, all the bulbs have the same voltage. But lamp C has the highest current.

Hence,

For Lamp C,

P = 5 x 60 W

= 300 W (the maximum).

(iv) The total current in the circuit = 3 + 4 + 5 + 3 A = 15 A

The Voltage = 60 V

V = IR and hence R = \(\frac VI\)

= \(\frac{60}{15}\) A

= 4 A

Answer 2

1: parallel combination

2: the most simple advanatage is that if anyone  of the lamp fails , circuit break due to any reason other lamps continue stop glowing .

there is identical potential difference at the end of each lamp .

3 :How does resistance  affect the brightness of the two bulbs?  The relationship between brightness and resistance of bulbs connected in parallel is inversely proportional. Therefore, the bulb with the lower resistance will shine brighter.

according to ohm's law V=IR

so, R=V/I

lamp1 , R= 60/3 = 20

lamp2 , R= 60/4 = 15

lamp3 , R= 60/5 = 12

lamp4 , R= 60/3 = 20

therefore lamp 3 which has less resistance will glow brightest .

4: we know that total resistance in parallel circuit is

1÷total resistance = 1/R1 +1/R2+1/R3+ ...........

thus 1/total resistance = 1/60/3 +  1/60/4 + 1/60/5 + 1/60/3

1/R total = 15/60

R total = 60/15

R total = 4 ohm