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 135 each. The battery supplies a voltage of 40 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 -0.148A
Bulb 2 -0.148A
Bulb 3 - 0 A
Bulb 4 - 0 A

(b) S3 is closed while S1 and S2 are open. What is the current through each bulb?
Bulb 1 -0.098 A
Bulb 2 - 0 A
Bulb 3 -0.098 A
Bulb 4 -0.098 A

(c) S2 and S3 are closed while S1 is open. What is the current through each bulb?
Bulb 1 -0.119 A
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 0 A
Bulb 4 ?

Explanation / Answer

a) is correct
b) is correct

c) equivalent to: B1, B2 in series with C
where C is B2 and B4 in parallel.
So total resistance: 135+135+ (135/2)=337.5
net current through the circuit: 40/337.5=0.119A
so B1=B3=0.119A
B2=B4=0.119/2=0.059A

d) equivalent to: B1 in series to C

where C is [A=3 and 4 in series] parallel with 2

so 270 parallel with 135 =270*135/(270+135)=90

so total R=225, therefore i=0.177A

so B1=0.177A

current then splits 2:1 due to resistance rations in C,

so B3=B4=0.177/3=0.059A

B2=0.177*2/3=0.118A

e)in the last one, yes B3=0A.

so its just B1 series with (B2 and B4)

Work this out on your own with an approach almost identical to part (c).

enjoy.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.