In the circuit shown below the 4 bulbs are identical and have a resistance of 19

Question

In the circuit shown below the 4 bulbs are identical and have a resistance of 190 ? each. The battery supplies a voltage of 20 V. S1. S2, and S3 represent switches.

(a) 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) 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) 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) S1 and S3 are closed while S2 is 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 Bulb 2 Bulb 3 Bulb 4

Explanation / Answer

a.) S1 and S2 closed, S3 open.

Due to closed S2, voltage across B3 = 0. Thus current throught B3 = 0.

Current through B4 = 0 (open circuit).

B1 and B2 come in series.

Thus same current of the value 20/(190+190) = 0.05263 amps flows through both B1 and B2.

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b.) S1 and S2 open means that B2 is open circuit while B1, B3, B4 are in series.

Current through B2 = 0;

Current through B1, B3, B4 = 20/(190*3) = 0.03509 amp.

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c.) B1 and B3 in series, while B2 and B4 in parallel.

Overall current = 20/(190*2 + 190/2) = 0.04210 amp.

Thus current through B1 and B3 = 0.04210 amp

while current through B2 and B4 = 0.04210/ 2 = 0.02105 amp.

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d.) Series combination of B3 and B4 is parallel to B2 and that overall is series with B1.

Overall resistance = 190 + (190 || 380) = 190 + 190*380/(190+380) = 316.6667 ohm

Thus overall current = 20/316.6667 = 0.06316 amp.

Thus current through B1 = 0.06316 amp.

B2 and B3||B4 share current in 2:1 ratio, with bigger chunk of current through B2 (due to lower resistance than series of B3 and B4)

B2 = 0.06316 / 3 = 0.021053 amp

B3 and B4 = 0.06316 * 2 / 3 = 0.042107 amp.

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e.) S1 and S2 closed means voltage across B3 = 0.

current through B3 = 0. (and thus branch can be eliminated from circuit, for analysis)

B2 and B4 parallel - combination in series with 1

Overall resistance = 190 + (190 ||190) = 190 + 95 = 285

Overall current = 20/285 = 0.07018 amp.

Thus current through B1 = 0.07018 amp.

Through B2 and B4 = 0.07018/2 = 0.03509 amp

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