Question

A bulb of rating 100 W, 220 volt is connected to a supply source of 110 V. If the power consumption required from the source is 100 W, then how many identical bulbs are required more to be connected in the circuit?
A. 3 more bulbs each in series with first
B. 4 more bulbs each in parallel with first
C. 3 more bulbs each in parallel with first
D. 4 more bulbs each in series with first

Answer 1

When bulb is connected with the power source it consumes some power. We are expecting the source to deliver some constant value of power so we should add a few bulbs to reach that value. Find the power consumed by one bulb first and then add bulbs accordingly to reach the required power value.

Formula used:
$P = \dfrac{{{V^2}}}{R}$

Complete step-by-step answer:
Let ‘R’ be the resistance of each bulb. When that bulb is connected to voltage source of voltage ‘V’ then the power consumed by that bulb will be $P = \dfrac{{{V^2}}}{R}$
Bulb rating is given as 100Watts and 220Volts. That means the bulb can be connected to the maximum voltage of 220 volts and can consume the maximum power of 100 watts. If we connect the bulb to a voltage source greater than that then the bulb will fuse.
When bulb of resistance ‘R’ is connected to the voltage source 220 volts then power consumed is 100 watts.
We have the power equation as $P = \dfrac{{{V^2}}}{R}$
$P = \dfrac{{{V^2}}}{R}$
$\Rightarrow 100 = \dfrac{{{{220}^2}}}{R}$ …eq 1
Now the same bulb is connected to the supply voltage source of 110volts. Then the power consumed by that bulb will be
$P = \dfrac{{{V^2}}}{R}$
$\Rightarrow P = \dfrac{{{{110}^2}}}{R}$
We need this power consumption to be 100 watts. So we are connecting some resistances to this existing resistance. Hence after connecting the effective resistance be ‘r’
Let us assume we connected the resistances in parallel hence voltage across every resistance will be 110 volts only and current will be divided. Hence power consumed by those resistors combined will be
$P = \dfrac{{{{110}^2}}}{r}$
This must be equal to 100 watts. Hence from equation 1 we get
$100 = \dfrac{{{{220}^2}}}{R}$
Now by substituting this equation 1 value in the effective power consumed by the resistors value we get
\eqalign{ & \Rightarrow P = \dfrac{{{{110}^2}}}{r} \cr & \Rightarrow 100 = \dfrac{{{{110}^2}}}{r} \cr & \Rightarrow \dfrac{{{{110}^2}}}{r} = \dfrac{{{{220}^2}}}{R} \cr & \Rightarrow \dfrac{{{{110}^2}}}{r} = \dfrac{{4 \times {{110}^2}}}{R} \cr & \Rightarrow r = \dfrac{R}{4} \cr}
Effective resistance is $\dfrac{R}{4}$ which will be possible only when we connect the three resistors in parallel to the existing resistor.

So, the correct answer is “Option C”.

Note: The other way to solve this problem is to find out the resistance of each bulb then power consumed by each bulb when connected to the given circuit and then finding the number of such bulbs needed. Since we have input voltage as a factor of rated voltage this question became easy to solve.