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
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.
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
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.
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.
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.
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.
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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