Four identical bulbs, each with a filament resistance of 144 ?, are connected to

Question

Four identical bulbs, each with a filament resistance of 144 ?, are connected to a 35-V battery as shown in the diagram below. S1, S2, and S3 represent switches.

(a) If S1 and S2 are closed while S3 is open, what is the current through each bulb?

bulb 1

bulb 2

bulb 3 bulb 4

(b) If S3 is closed while S1 and S2 are open, what is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

(c) If S2 and S3 are closed while S1 is open, what is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

(d) If S1 and S3 are closed while S2 is open, what is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

Explanation / Answer

a)
Here current will flow through:
battery ---> S1-->bulb 2 --> bulb 1 ---> battery
current through 1 = current through 2 = V / (R+R)
=35/(144+144)
=0.122 A

current thorugh 3=0
current thorugh 4=0

b)
current will ot be able to flow as circuit is not complete
current through 1 = current through 2 = current through 3 = current through 4 = 0

c)
2 and 4 are in parallel which are in turn in series with 1 and 3
Rnet = (R2*R4)/(R2+R4) + R1+R3
=72 + 144 + 144
=360 ohm

current through 3 =current through 1 = 35/260 = 0.0972 A
current through 2= current through 4 = 0.0972/2 = 0.0486 A

d)
2 is in parallel with 3 and 4 which in turn is in series with 2
Rnet = (R2||(R3+R4) )+R1
= (144||288) + 144
= 240 ohm

I thorugh 1= 35/240 = 0.146 A
V drop across 1 = 0.146*R1 = 0.146*144 =21 V
V drop across 2 = 35-21 = 14 V

I thorugh 2= 14/144 = 0.0972 A

I though 3 = I through 4 = 0.146 - 0.0972 = 0.0488 A

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