Answer / nitin narwal

bulbs or any domestic electrical equipments are designd for
constant voltage,thats why we have parallel connections at
home.and r designd for the formula given below intentionally
we know that p=v^2/r
now p is inversely proportional to r
now we can easily tell 50w bulb will have less resistance
than 30w bulb
now in series, current in both will same,,,,,,so by formula
p=i^2R,,,,,
we can say 30w bulb will glow brighter

Answer / mir razin bashar

we know that power is inversely proportional to resistance.
p=(v^2)/R where V is constant
i is not constant because current is supplied
by total constant source voltage.
but current are vary by adding resistance

so
the bulb having more power has has less resistance
less power more resistance

example:
50w bulb will have less resistance than 30w bulb

brightness:
series:
50w and 30 w are connected.
in series current against resistance is fixed.(totla source voltege is const)
so energy rejected
H(50w)=(i^2)*R*t
H(30w)=(i^2)*r*t
r=resistance of 30w bulb
R=resistance of 50w bulb
we can tell according to the above description
r>R
=>r*t>R*t
=>(i^2)*r*t>(i^2)*R*t
=>H(30w)>H(50w)
30w is more brighter than 50w
parallel:
50w and 30 w are connected.
in parallel voltage against resistance is fixed.(also totla source voltege is const)
so energy rejected
H(50w)=(v^2)*t/R
H(30w)=(v^2)*t/r
r=resistance of 30w bulb
R=resistance of 50w bulb
we can tell according to the above description
r>R
=>(1/r)<(1/R)
=>((v^2)*t/r)<((v^2)*t/R)
=>H(30w)<H(50w)
50w is more brighter than 30w

so in house as parallel circuit is used so we brought 100w than 50w to get more light
but if series circuit is used so we brought 50w than 100w to get more light

Answer / ahir

AS we know for series ckt p=i^2/r...

so, in which buld have low resistance will blow more.....

and 50W BULB have less r..

so , it blow brightly....

Answer / naresh.

p=v*i1 v=i1*r1
50=230*i1 r1=230/0.217
i1=(0.217A) r1=1059 ohms
p=v*i2 v=i2*r2
30=230*i2 r2=230/0.130
i2=(0.130A) r2=1769 ohms

obvious that one which is drawing more current has least
resistance path as it in series the current takes the least
resistance path and we can also see that 30w bulb has high
resistance so less curentt flows through it so it is less
brighter than 50w

conclusion:
so 50w bulb glows brighter

Answer / md. zakir hosen

We know that, P=VI
here,
P=50+30=80w
v=230
I=?
its very simple to calculate and see that the current is
flowing throw the circuit is suitable for the LOW power
equipments as a result 30W bulb will be the brighter.

thanking u
z_hosen@yahoo.com

Answer / arshad noor

As we Know that current in series flow according to Lowest
value of resistance therefore 30W will glow bright since it
has low resistance compared to 50W

Answer / adel

current pass in two bulb= p/v=80/230=0.35 A
voltage on bulb 50 W= p/i=50/0.35=142 V
voltage on bulb 30W= p/i=30/0.35 =85.7 V
hence bulb 50w is bright

Answer / felixraj

totally 80w ,i=p/v 80/230=0.34A vl droop in 50w is more
(brighter)

Answer / manesh.k

P=VI,
So this combined 80w load will draw approx. 0.34 amp from
the 230 v supply

So when the same current flow through the bulbs,the highest
resistance one,ie 50w bulb will glow bright.

Answer / senil

the lamp which is connected near to phase that lamp is
glown bright........

How does flights get the neutral(grounding)

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