Question

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.​
plz answer me fast its urgent today is my pre board​

Answer 1

Explanation:

The lamps are in parallel.

Advantages: If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel. The lamp with the highest power will glow the brightest. P=VI 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 Watt = 300 W. (the maximum). The total current in the circuit = 3+4+5+3 A = 15A The Voltage = 60V V=IR and hence R =V/I = 60/15 A = 4A

Answer 2

Explanation:

i. Parallel combination is best suited way of arranging lamps here.

ii.

• In case of failure of any of the lamps, the other lamps will continue to work since the circuit will not be affected.
• Parallel combination reduces overall resistance of the circuit.
• parallel combination has equal distribution of voltage.

iii. Brightness of a lamp depends upon the power it consumes. Hence, the less the resistence of the lamp will be the more will it consume power and the more brighter it will be.

iv. Total resistence in this circuit will be;

• If A has resistence = R1
• B has resistence = R2
• C has resistence = R3
• D has resistence = R4

then, 1 / R eff = 1//R1 + 1/R2 + 1//R3 + 1/R4. (R eff = Total Resistance)

R eff = R1.R2.R3.R4 / (R1 + R2 + R3 + R4).

That's all.