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if two bulb of 50w &30w apply in series having 230v the
which bulb glow bright ,why
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Answer Posted / 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
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