
A light bulb is rated 100 Watts for 220 Volts AC supply of 50 Hertz. Calculate
(i) The resistance of the bulb
(ii) The rms current through the bulb
Answer
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Hint: In this question, we need to determine the resistance of the bulb as well as the rms value of the current flowing through the bulb. For this, we will use the relation between the power, voltage and current in an AC supply.
Complete step by step answer:(i)The ratio of the square of the rms value of the voltage across the element and the resistance of the element results in the power consumed (for passive elements) or delivered (for active elements) by the element. Mathematically, $P = \dfrac{{{V^2}_{rms}}}{R}$ where${V_{rms}}$ is the rms value of the voltage and R is the resistance of the element.
According to the question, the power rating of the bulb is 100 watts, and the rms value of the voltage across the bulb is 220 volts.
So, substitute P=100 watts and ${V_{rms}}$=220 volts in the formula $P = \dfrac{{{V^2}_{rms}}}{R}$ to determine the resistance of the bulb.
$
P = \dfrac{{{V^2}_{rms}}}{R} \\
100 = \dfrac{{{{\left( {220} \right)}^2}}}{R} \\
R = \dfrac{{220 \times 220}}{{100}} \\
= 22 \times 22 \\
= 484\Omega \\
$
Hence, the resistance of the bulb is 484 ohms.
(ii)The product of the rms value of the current through an element and the resistance of the element results in the rms value of the voltage across the element. Mathematically, ${V_{rms}} = {I_{rms}}R$ where, ${V_{rms}}$ and ${I_{rms}}$ are the rms value of the voltage and the current respectively and R is the resistance of the element.
So, substitute ${V_{rms}}$=220 volts and R=484 ohms in the formula ${V_{rms}} = {I_{rms}}R$ to determine the rms value of the current flowing through the bulb.
$
{V_{rms}} = {I_{rms}}R \\
220 = {I_{rms}} \times 484 \\
{I_{rms}} = \dfrac{{484}}{{220}} \\
= 2.2{\text{ Amperes}} \\
$
Hence, the value of the rms current flowing through the bulb is 2.2 Amperes.
Note:It is worth noting here that the value of the voltage given in the question for the AC source should always be taken as the rms value only. Rms value is the root mean square value of the source voltage or current. It is the measure of the heat transfer between the elements.
Complete step by step answer:(i)The ratio of the square of the rms value of the voltage across the element and the resistance of the element results in the power consumed (for passive elements) or delivered (for active elements) by the element. Mathematically, $P = \dfrac{{{V^2}_{rms}}}{R}$ where${V_{rms}}$ is the rms value of the voltage and R is the resistance of the element.
According to the question, the power rating of the bulb is 100 watts, and the rms value of the voltage across the bulb is 220 volts.
So, substitute P=100 watts and ${V_{rms}}$=220 volts in the formula $P = \dfrac{{{V^2}_{rms}}}{R}$ to determine the resistance of the bulb.
$
P = \dfrac{{{V^2}_{rms}}}{R} \\
100 = \dfrac{{{{\left( {220} \right)}^2}}}{R} \\
R = \dfrac{{220 \times 220}}{{100}} \\
= 22 \times 22 \\
= 484\Omega \\
$
Hence, the resistance of the bulb is 484 ohms.
(ii)The product of the rms value of the current through an element and the resistance of the element results in the rms value of the voltage across the element. Mathematically, ${V_{rms}} = {I_{rms}}R$ where, ${V_{rms}}$ and ${I_{rms}}$ are the rms value of the voltage and the current respectively and R is the resistance of the element.
So, substitute ${V_{rms}}$=220 volts and R=484 ohms in the formula ${V_{rms}} = {I_{rms}}R$ to determine the rms value of the current flowing through the bulb.
$
{V_{rms}} = {I_{rms}}R \\
220 = {I_{rms}} \times 484 \\
{I_{rms}} = \dfrac{{484}}{{220}} \\
= 2.2{\text{ Amperes}} \\
$
Hence, the value of the rms current flowing through the bulb is 2.2 Amperes.
Note:It is worth noting here that the value of the voltage given in the question for the AC source should always be taken as the rms value only. Rms value is the root mean square value of the source voltage or current. It is the measure of the heat transfer between the elements.
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