Four identical bulbs, each with a filament resistance of 135 , are connected to
ID: 1584100 • Letter: F
Question
Four identical bulbs, each with a filament resistance of 135 , are connected to a 30-V battery as shown in the diagram below. S1, S2, and S3 represent switches.
In the circuit there are 4 circles, labeled 1, 2, 3, and 4. There are three switches, labeled S_1, S_2, and S_3. There is one battery, labeled V. Beginning at the positive terminal of the battery, the wire splits. The left branch has switch S_1 and the right branch has circle 3. The branches are bridged by a wire with switch S_2. After the wires are connected by S_2, the left and right branches continue. The left branch has circle 2, and the right branch has circle 4 and then switch S_3. The left and right branches then reconnect, pass through circle 1, and then end at the negative terminal of the battery.
(a) If S1 and S2 are closed while S3 is open, what is the current through each bulb?
bulb 1 A
bulb 2 A
bulb 3 A
bulb 4 A
(b) If S3 is closed while S1 and S2 are open, what is the current through each bulb?
bulb 1 A
bulb 2 A
bulb 3 A
bulb 4 A
(c) If S2 and S3 are closed while S1 is open, what is the current through each bulb?
bulb 1 A
bulb 2 A
bulb 3 A
bulb 4 A
(d) If S1 and S3 are closed while S2 is open, what is the current through each bulb?
bulb 1 A
bulb 2 A
bulb 3 A
bulb 4 A
tical bulbs, each with a filament resistance of 135 n, are connected to a 30-V S1 ) If S, and S, are closed while S, is open, what is the current through each bulk ulb 1 Wha: happens ro bul- 2 hen s. s closec? If s: s oen. is there a path hulh 2Explanation / Answer
a) Rnet = 135 + 135 = 270 ohms
Current through bulb 1, I1 = V/Rnet = 30/270 = 0.111 A
Current through bulb 2, I2 = 0.111 A
Current through bulb 3, I3 = 0 A
Current through bulb 4, I4 = 0 A
b) Rnet = 135 + 135 + 135 = 405 ohms
Current through bulb 1, I1 = V/Rnet = 30/405 = 0.074 A
Current through bulb 2, I2 = 0 A
Current through bulb 3, I3 = 0.074 A
Current through bulb 4, I4 = 0.074 A
c) Rnet = 135 + 135/2 + 135 = 337.5 ohms
Current through bulb 1, I1 = V/Rnet = 30/337.5 = 0.089 A
Current through bulb 2, I2 = 0.89/2 = 0.044 A
Current through bulb 3, I3 = 0.089 A
Current through bulb 4, I4 = 0.89/2 = 0.044 A
d) Rnet = 135 + 270*135/(270 + 135) = 225 ohms
Current through bulb 1, I1 = V/Rnet = 30/225 = 0.133 A
Current through bulb 2, I2 = 0.133*270/(135 + 270) = = 0.089 A
Current through bulb 3, I3 = 0.133*135/(135 + 270) = 0.044 A
Current through bulb 4, I4 = 0.044 A
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