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Solve it, In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

 In a large building, there are 15 bulbs of   40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW.  The voltage of the electric mains is 220 V.  The minimum capacity of the main fuse of the building will be :

 

  • Option 1)

    8 A

  • Option 2)

    10 A

  • Option 3)

    12 A

  • Option 4)

    14 A

 
Answers (2)
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N neha

As we have learned

In parallel Grouping -

\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\cdot \cdot \cdot +\frac{1}{R_{n}}

-

 

 Resistance of each bulb of 40 W = \frac{v^2 }{P } = \frac{220 ^2 }{40 }= 1210 \Omega

 

Resistance of 15 bulbs = 1210/15 = 242/3 \Omega

Resistance of 100 W bulb = \frac{v^2 }{P }= \frac{220 ^2 }{100} = 484 \Omega

 

Resistance of 5 such bulbs = 485/5 \Omega

Resistance of fan =  \frac{v^2 }{P }= \frac{220 ^2 }{80}   =  605 \Omega

Resistance of 5 such fan  = 605 /5 \Omega

= 121 \Omega

Resistance of heater = \frac{220 ^2 }{1000} = 242 /5\Omega

 Equivalent resistance 

\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}+ \frac{1}{R_4}\\ = \frac{3}{242} + \frac{5}{484}+ \frac{5}{605}+\frac{5}{242} \\

= \frac{6+5+4+10}{484} = \frac{25 }{484}

R = 484 / 25 

I \geq \frac{V}{}R = \frac{220 }{\frac{484}{25}} = \frac{125 }{11} = 11.36 A

I^{min } = 12 A

 

 

 

 

 

 

 

 


Option 1)

8 A

Option 2)

10 A

Option 3)

12 A

Option 4)

14 A

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