Question

On a bulb is written $220\;V$ and $60\;watt$. Find the resistance of the bulb and the value of the current flowing through it.
A. $806.66\;\Omega$ or $0.27\;A$
B. $500\;\Omega$ or $2\;A$
C. $200\;\Omega$ or $4\;A$
D. $100\;\Omega$ or $1\;A$

Answer 1

Recall that electric power can be defined as the work done in moving charges to produce a current under a potential difference. Use this to determine an expression correlating power, voltage and current, to find the current flowing through the bulb. Then, use Ohm’s law to express current in terms of voltage and resistance and plug the resultant back into the previous equation to obtain the appropriate resistance of the bulb.

Formula Used:
Power $P=\dfrac{V^2}{R} = IV$

Complete Solution:
Let us begin by first understanding the concept of electric power and Ohm’s Law, following which we shall derive expressions to find the resistance of the bulb and the current flowing through it, using the parameters given to us.

Electric power represents the rate at which electrical energy is transferred through an electric component. Like mechanical power, it is defined as the rate of doing work and is measured in watts (W), which is equivalent to one joule (J) per second (s). For electrical energy to move electrons or charges (Q) and produce a current in a circuit, there must exist a potential difference (V) across the ends of the circuit, i.e.,
Work done $W = QV$

Therefore, power, which is nothing but the rate of doing work is given as:
$P = \dfrac{W}{t} = \dfrac{QV}{t}$

However, we know that the charge flowing through a region of a circuit per unit time is defined as the electric current, i.e.,
$P = \dfrac{Q}{t}.V = \dfrac{I}{V}$

Now, in the context of the question. We canl use the above expression to find the current flowing through the bulb, given that voltage $V = 220\;V$ and power $P=60\;W$:
$I = \dfrac{P}{V} = \dfrac{60}{220} = 0.27\;A$

Next, we have Ohm’s Law which states that the voltage across a conductor is directly proportional to the current flowing through it:
$V\propto I \Rightarrow V=RI \Rightarrow I = \dfrac{V}{R}$, where R is the resistance offered by the conductor.

Substituting this in our power equation, we get:
$P=IV =\left(\dfrac{V}{R}\right).V = \dfrac{V^2}{R} \Rightarrow R= \dfrac{V^2}{P}$
$\Rightarrow R = \dfrac{220^2}{60} = \dfrac{48400}{60}=806.66\;\Omega$

Therefore, the correct choice would be A. $806.66\;\Omega$ or $0.27\;A$

Note:
An alternate way to understand the above problem would be to use Joule’s Law of heating, which describes the rate at which the resistance of an electrical component converts electric energy into heat or light energy, as is in the case of a bulb. The heat or light energy is generated due to the flow of charges through the electrical component, and is quantitatively describes as:
Heat generated $Q = I^2 Rt$, which can be rewritten as: $\dfrac{Q}{t} = I^2 R \Rightarrow P = I^2R = I^2.\dfrac{V}{I} \Rightarrow P = IV$, which is the same as what we obtained anyways.