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A circuit has a fuse of 5A.find the maximum number of 100W 220V lamp that can be used in a circuit.

Given Power - 100W

Given Voltage - 220V

We know

P = V*I where P = power, V = voltage and I = current

Applying this formula we get

100 = 220 * I

=> I = 100/220 = 0.4545454545..........A = 0.45A approx.

Now current required to run one lamp of 100W 220V = 0.45

Total current supply = 5A

Therefore, no. of appliances which can be used at one time - 5/0.45 = 11.1111111......... = 11.11 

Since appliances cannot be taken in points, the number of appliances will be 11.

Note: It wont matter whether the lamps are connected in series or parallel. The current drawn by each lamp will be the same, i.e. the current flowing through each lamp will be the same as in series connection, and in parallel connection, since V is always the same, and the resistances of all the lamps is also the same, the current will be same. 

Hope this helps!!

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View Full Answer

we know, 

P = V*I --> I =P/V

    =100/220

=5/11 A

so the current needed for 1 lamp= 5/11 A

maximum allowable current = 5A

no. of lamps = 5/5/11

= 5*11/5

=11 lamps....

tell me if there is any mistake...

  • 36
Is this correct? 
Take the no. of appliances as 'x' 
we know, P= VI 
so total voltage supply, 220V 
current= 5A
; 100*x = 220*5 
, x = 1100/100
x = 11
pls tell if my approach to the question is correct.

 
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Suppose x number of such bulbs can be used Power of one bulb = 100 watt Power of x bulbs (p) = 100 X x watts Potential difference (V) = 220 volts I = 5 ampere We know that P = VI 100 X x = 220 x 5 x = 220 x 5 /100 = 11 bulbs.
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I=5A V=
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This picture contains the best answer.. I hope it will be helpful...

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P=VI

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Sove as fast as you can plz.....

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Help me plz solve it as fast as u can
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V/p × I where v= voltage p= power and I = current Therefore, 220/100×5 =11 ans
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You are such an
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F#@$#@*a@*#$*@#
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what is this mahn
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hi
 
  • -4
Suppose x number of such bulbs can be used 

Power of one bulb = 100 watt 

Power of x bulbs (p) = 100 X x watts 

Potential difference (V) = 220 volts 

I = 5 ampere 

We know that P = VI 

100 X x = 220 x 5 

x = 220 x 5 /100 = 11 bulbs.
 
  • -2
Suppose x number of such bulbs can be used

Power of one bulb = 100 watt

Power of x bulbs (p) = 100 X x watts

Potential difference (V) = 220 volts

I = 5 ampere

We know that P = VI

100 X x = 220 x 5

x = 220 x 5 /100 = 11 bulbs.
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Power 100* x = 220*5 x=11 Hope this is right
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Let the maximum number of lamp be x Power of x lamp is p=100x Then, using formula p=VI And get the answer....
  • -2
Suppose x number of such bulbs can be used 

Power of one bulb = 100 watt 

Power of x bulbs (p) = 100 X x watts 

Potential difference (V) = 220 volts 

I = 5 ampere 

We know that P = VI 

100 X x = 220 x 5 

x = 220 x 5 /100 = 11 bulbs
 
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