6 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.
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Answer:
(i) Parallel combination
(ii) #There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
#The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
(iii)A and D bcoz of high resistance
hopefully I am right
Answer:
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