Four identical bulbs, each with a filament resistance of 151 ohm, are connected
ID: 1491869 • Letter: F
Question
Four identical bulbs, each with a filament resistance of 151 ohm, are connected to a 50-V battery as shown in the diagram below. S_1, S_2, and S_3 represent switches. If S_1 and S_2 are closed while S_3 is open, what is the current through each bulb? If S_3 is closed while S_1 and S_2 are open, what is the current through each bulb? If S_2 and S_3 are closed while S_1 is open, what is the current through each bulb? If S_1 and S_3 are closed while S_2 is open, what is the current through each bulb?Explanation / Answer
(a) In this case, there is a short circuit from Bulb 1 to Bulb 2, so R-equiv is 302 ohms,
and the current through each of Bulbs 1 and 2 is = 50/302 =0.1655 amps. No current through 3 and 4.
(b) In this case, Bulbs 1,3,4 are in series, R-equiv is 453 ohms; current through 1,3,4 is 0.1103 amps.
No current through Bulb 2.
(c) In this case, Bulbs 1 and 3 are in series with the parallel-resistance combination of Bulbs 2 and 4.
R-equiv for the whole circuit is 2.5 x (151 ohms), and the current through Bulbs 1 and 3 is 0.132 amps; current is split through Bulbs 2 and 4, 0.066 amps each.
(d) In this case, Bulb 1 is in series with a parallel combination whose two branches are (i) Bulb 2 and (ii) Bulbs 3 and 4 in series. R-equiv for the whole circuit is
151 + 1/(1/151+1/302) = 251.666 ohms.
Current through Bulb 1 is 50 V / 251.667 ohms = 0.198 amps.
The parallel branches split this current, with twice as much of it going through Bulb 2 as through the 3-4 series combo. So the current in Bulb 2 is 0.132 amps, and the current through Bulbs 3 and 4 is 0.066 amps.
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