A bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also, find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change.

(i) Given:

Power of the bulb, $P_B=40W$

Voltage, $V=220V$

To find: Current drawn by the bulb, $I$.

Solution:

We know that electric power is given as-

$P=V\times I$

Therefore, 

$I=\frac {P}{V}$

Substituting the required value, we get-

$I=\frac {40}{220}$

$I=0.18A$

Thus, the current drawn by the bulb is 0.18A.

We know that resistance is given as-

$R=\frac {V^2}{P}$

Putting the required value we get-

$R=\frac {220^2}{40}$

$R=\frac {220\times 220}{40}$

$R=1210\Omega$

Thus, the resistance of the 40 W bulb is 1210 ohm.

Now, the given bulb is replaced by a bulb of rating 25 W; 220 V.

Given:

Power of the bulb, $P_B=25W$

Voltage, $V=220V$

To find: Current drawn by the bulb, $I$, and resistance, $R$.

Solution:

We know that electric power is given as-

$P=V\times I$

Therefore, 

$I=\frac {P}{V}$

Substituting the required value, we get-

$I=\frac {25}{220}$

$I=0.113A$

Thus, the current drawn by the bulb is 0.113A.

We know that resistance is given as-

$R=\frac {V^2}{P}$

Putting the required value we get-

$R=\frac {220^2}{25}$

$R=\frac {220\times 220}{25}$

$R=1936\Omega$

Thus, the resistance of the 25 W bulb is 1936 ohm.

Yes, there is a change in current and resistance.

$\Rightarrow change\ in\ current=0.18-0.1136=0.0664\ A$

$\Rightarrow change\ in\ resistance=1936-1210=726\Omega$

Hence, from the above justification, we can see that current decreases and resistance increases when we use a 25Watt bulb in place of a 40Watt.