Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 12)
12.
Five light bulbs are connected in parallel across 110 V. Each bulb is rated at 200 W. The current through each bulb is approximately
2.2 A
137 mA
1.8 A
9.09 A
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Nilesh lale said:   1 decade ago
In this voltage drop across each resistance is 110 we because they are in parallel. So by using equation P=VI, We get I=P/V.

That means I= 200/110= 1. 8 A.

Ujjwal roy said:   1 decade ago
Total watts will be 1000 W AS 200*5=1000 i.e.1000/110=9.09A

Arundeep singh said:   1 decade ago
Right one is 200 w x 5 = 1000 w
v = 110 v
So total current 1000/110 = 9.0909
There are 5 bulb (resister) of same rating 200 w
So 9.0909/5 = 1.818

Acbob said:   1 decade ago
@Nilesh's answer is simlpe, please others don't add complex procedures.

SWETA said:   1 decade ago
W = VI.
W/V = I.

WHERE W = 200, V = 110.
200/110 = I.
1.81 = I.

Kailash said:   9 years ago
W = v^2 * r.

R = 110^2/200 = 60.5.
Req =60.5/5 = 12.1.

I = 110/12.1 = 9.09.

Aditya kumar said:   9 years ago
There are 5 bulbs here. Each bulb has 200 w.

So 200 * 5 = 1000 w (total power).

That is, 1000/110 = 9.0909.

So each bulb is = 9.0909/5 = 1.81 A.
(1)

Sushil said:   5 years ago
Can we add wattages when we considering they are connected in parallel?

Fakhur said:   3 years ago
I = P/V= 200/110 = 1.8181 for each bulb,
Assuming the power supply is well-regulated - doesn't sag under load, the total is
1.8181 * 5 = 9.09A for the total current.

Arpan Roy said:   1 year ago
The right answer should be 9.09A.

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