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The lamps are in parallel. <br> Advantages: <br> 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. <br> The lamp with the highest power will glow the brightest. <br> P=VI <br> 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). <br> The total current in the circuit = 3+ 4 + 5 + 3 = 15 A <br> The Voltage = 60 V <br> V = IR and hence R = V/I <br> = 60/15 A = 4A <br> V= IR and hence R = V/I <br> = 60/15 A = 4A <br>
i] The given circuit is connected in parallel combination [ this is because they are not connected end to end]
ii] 1) In parallel combination each appliance gets the full voltage.
2) If one appliance is switched on/off others are not affected.
3) The parallel circuit divides the current through the appliances. Each appliance gets a proper current depending on its resistance.
iii] V = IR; V = 60
I = 3A
R = 60/3 = 20 ohm- A
R = 60/4 = 15 ohm-B
R= 60/5 = 12 ohm-C
R= 60/3 = 20 ohm-D
As resistance is inversely proportional to current, The one with the least resistance will glow brighter. Hence bulb C will glow brighter
iv] 1/Rp = 1/R1 + 1/R2 + 1/R3 + I/R4
= 1/20 + 1/15 + 1/12 + 1/20
= 3+4+5+3/60
=15/60
= 1/4
Rp = 4 ohm
Total resistance = 4ohm