Electrical Machines and Circuits

Total power P = V * I = 220 V * 5 A = 1100 W. Since each bulb is 20 W, the number of bulbs n = 1100 / 20 = 55.

In a DC circuit, the soft iron core increases the inductance of the coil. However, since the current is steady DC, the inductive reactance is zero, so the intensity of the bulb remains unchanged.

The electric bulb has a filament called tungsten when electricity passes through this filament, it heats up and glows. 

Resistance$,\ R= \dfrac{V^2}{P} =\dfrac{(220)^2}{P}$. Since $P$ of each appliance is the same, hence $R$ is same for all the three appliances.

For bulbs :

$Energy\,\,=\,power\times time$

Total hours in 30 days $5\times 30=150hours$

So, $ energy=0.1\times 150=15J $

 $  $Power of 4 bulbs

 $ =4\times 15=60kW $

 $  $For LED:

Total hours in 30 days $=150hours$

$E=0.008\times 150=1.2J$

So, power of 4 LEDs $=1.2\times 4=4.8hours$

Saved energy 

  $ =\,\,60-4.8 $

 $ =55.2units $

Power P is proportional to V^2. If V decreases by 2.5%, V_new = 0.975 V_old. P_new is proportional to (0.975)^2 = 0.9506. The decrease is approximately 5%.

An ohmmeter measures resistance at low voltage/current. A tungsten bulb's resistance increases significantly with temperature. When in use, the filament is hot, so its operating resistance is much higher than its cold resistance measured by an ohmmeter.

Heat lost per second per unit surface area of fuse wire is
$H = \frac{I^2 \rho }{2 \pi ^2 r^3}$
$\Rightarrow  I^2 \propto r^3$
$\Rightarrow  I \propto r^{\frac{3}{2}}$

$V = 200   V,   P = 750   \omega$.
$P = VI ,   750 = 200 \times I$
$I = 3.75   A$.

$E = \dfrac{{V}^{2}t}{R}$
  $= \dfrac{{\left(200\right)}^{2} \times 10}{480}  =  1008.3  J$

Resistance of the bulb $R = \dfrac{V^2}{P}$

Why is the switch for any electrical appliance always fitted on the live wire.

Answer is C.

As the three bulbs are connected in parallel, the voltage applied across them would be the same, lets say its V.
Now, due to the fact that the power dissipated by the three bulbs is different the current flowing through them would be different and thus they will have different brightness.
Now, the brightness of a bulb is directly proportional to the power it uses or the amount of current it draws. 
Thus, in this case the brightness of bulb B (60 W) will be more than that of the bulb A (40 W) but less than that of the bulb C (100 W). That is, C has the maximum brightness and the bulb has has the least brightness. Bulb B in between.
Hence, the correct statement is option C.

Resistance of Ist bulb $={ R } _{ 1 }=\cfrac { { V }^{ 2 } }{ { P } _{ 1 } } $
Resistance of IInd bulb $={ R } _{ 2 }=\cfrac { { V }^{ 2 } }{ { P } _{ 2 } } $
when both bulbs are connected in series
$\quad { R } _{ eq }={ V }^{ 2 }\left[ \cfrac { 1 }{ { P } _{ 1 } } +\cfrac { 1 }{ { P } _{ 2 } }  \right] =\cfrac { { V }^{ 2 }\left( { P } _{ 1 }+{ P } _{ 2 } \right)  }{ { P } _{ 1 }{ P } _{ 2 } } $
Hence, power consumed $P=\cfrac { { V }^{ 2 } }{ R } =\cfrac { { V }^{ 2 } }{ { V }^{ 2 }\left( \cfrac { { P } _{ 1 }+{ P } _{ 2 } }{ { P } _{ 1 }{ P } _{ 2 } }  \right)  } =\cfrac { { P } _{ 1 }{ P } _{ 2 } }{ { P } _{ 1 }+{ P } _{ 2 } } \quad $