Q

# 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

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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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