In the circuit shown below the 4 bulbs are identical and have a resistance of 10

Question

In the circuit shown below the 4 bulbs are identical and have a resistance of 105 each. The battery supplies a voltage of 25 V. S1, S2, and S3 represent switches.

(a) S1 and S2 are closed while S3 is open. What is the current through each bulb?

(b) S3 is closed while S1 and S2 are open. What is the current through each bulb?

(c) S2 and S3 are closed while S1 is open. What is the current through each bulb?

(d) S1 and S3 are closed while S2 is open. What is the current through each bulb?

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

Explanation / Answer

a) bulb 3 is short circuited

so only 1 and 2 are considered

Reff = 105+105 = 210 ohm

i = V/Reff = 25/210 = 0.119 A

so bulb 1 = 0.119 A

bulb 2 = 0.119 A

bulb 3 = 0 A

bulb 4 = 0A

b) bulb 1,3, 4 are in series

Reff = 105+105+105 = 315 ohm

i = V/Reff 25/315 = 0.08 A

bulb 1 = 0.08 A

bulb 2 = 0 A

bulb 3 = 0.08 A

bulb 4 = 0.08 A

c) bulb 2 paralle with 4 and the combination in series with 1 and 3

Reff = 105+105 + 105/2 = 262.5 ohm

i = V/Reff = 25/262.5 = 0.095 A

bulb 1 = 0.095 A

bulb 2 = 0.095/2 = 0.0476 A

bulb 3 = 0.095 A

bulb 4 = 0.095/2 = 0.0476 A

d) 3 series with 4 and the combination parallel with 2 and this whole combination series with bulb1

Reff = 105 + [105*(105+105)/ (105+105+105)] = 175 ohm

i = V/Reff = 25/175 = 0.143 A

bulb 1 = 0.143 A

bulb 2 = (2/3)*0.143 = 0.095 A

bulb 3 = (1/3)*0.143 = 0.0476 A

bulb 4 = (2/3)*0.143 = 0.0476 A

e) 3 is short circuited

bulb 1 in series with combination of buld 2 and 4 in parallel

Reff = 105 + 105/2 = 157.5 ohm

i = V/Reff = 25/157.5 = 0.1587 = 0.16 A

bulb 1 = 0.16 A

bulb 2 = 0.1587/2 = 0.08 A

bulb 3 = 0 A

bulb 4 = 0.16/2 = 0.08 A

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