In the circuit shown below the 4 bulbs are identical and have a resistance of 13

Question

In the circuit shown below the 4 bulbs are identical and have a resistance of 130 ? each. The battery supplies a voltage of 20 V. S1, S2, and S3 represent switches.

(a) S1and S2are closed while S3is open. What is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

b) S3is closed while S1and S2are open. What is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

(c) S2and S3are closed while S1is open. What is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

(d) S1and S3are closed while S2is open. What is the current through each bulb?

bulb 1

bulb 2

bulb 3

bulb 4

(e) All three switches are closed. What is the current through each bulb?

bulb 1

bulb2

bulb 3

bulb 4

Explanation / Answer

a) as S3 is open , so current through bulb 4 = 0

as S1 and S2 both are closed , so bulb 3 is short cicuited , so current through bulb 3 = 0

so, entire current pass through bulb 1 and 2

Bulb 1 = Bulb 2 = 20 / 260 = 0.077 A

bulb 3 = bulb 4 = 0

b) In this case bulb 2 is disconnected and entire current pass through rest of three bulbs

bulb1 = bulb3 = bulb 4 = 20 / 390 = 0.513 A

bulb 2 = 0

c) In this case bulb 2 and bulb 4 will be in parallel so

resistance across parallel = 130/2 = 65 ohm

total resistance of circuit = 260 + 65 = 325 ohm

current = 20/325 = 0.06154 A

bulb 1 = bulb 3 = 0.06154

bulb 2 = bulb 4 = 0.06154/2 = 0.03077 A

d)

Bulb 3 and bulb 4 in series

resistance across series = 130 + 130 = 260 ohm

the series combination of bulb 3 and 4 is in parallel witth bulb 2

so, resistance across parallel = 260*130/(130 + 260 ) = 86.67 ohm

total resistance of circuit = 130 + 86.67 = 216.67 ohm

current = 20/216.67 = 0.09231 A

bulb 1 = 0.09231

bulb 3 = bulb 4 = (1/3) * 0.09231 = 0.03077 A

bulb 2 = (2/3) * 0.09231 = 0.06154 A

e) bulb 3 will be short circuited , so current through bulb 3 = 0

bulb 2 and bulb 4 will be in parallel so

resistance across parallel = 130/2 = 65 ohm

total resistance of circuit = 130 + 65 = 195 ohm

current = 20/195 = 0.10256 A

bulb 1 = 0.10256 A

bulb 2 = bulb 4 = 0.10256 / 2 = 0.05128 A

bulb 3 = 0

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