Question

An electric bulb rated for 500 watt at 100 volt is used in a circuit having a 200 volt supply. Resistance R that must be connected in series with the bulb so that the bulb delivers 500 watts is
 A. \[100\;\Omega \]
B. \[20\;\Omega \]
C. \[60\;\Omega \]
D. \[80\;\Omega \]

Answer 1

Hint: The basic formula for the power dissipation through the conductor carrying some current, for the given voltage supply is applied. Moreover, the concept of Ohm’s law is also used.

Complete step by step answer:
The power rating for the electric bulb is, \[P = 500\;{\rm{W}}\].
The voltage supply for the electric bulb is, \[{V_1} = 100\;{\rm{V}}\].
 The voltage supply in a circuit is, \[{V_2} = 200\;{\rm{V}}\].
The resistance is given as,
\[\begin{array}{l}
R = \dfrac{{V_1^2}}{P}\\
R = \dfrac{{{{\left( {100\;{\rm{V}}} \right)}^2}}}{{500}}\\
R = 20\;\Omega
\end{array}\]
For the power rating of 500 Watts,
 \[\begin{array}{l}
P = {I^2} \times R\\
500\;{\rm{W}} = {I^2} \times 20\;\Omega \\
I = 5\;{\rm{A}}
\end{array}\]
Now, for the above value of current in the series connection, the resistance is calculated using the Kirchhoff’s Voltage Rule as,
 \[\begin{array}{l}
V_2 = I\left( {R + R'} \right)\\
200\;{\rm{V}} = 5\;{\rm{A}}\left( {20\;\Omega + R'} \right)\\
R' = 20\;\Omega
\end{array}\]
Therefore, the required value of resistance in the series connection is \[20\;\Omega \]

So, the correct answer is “Option B”.

Note:
To resolve the given problem, the fundamental relations of power supply in an electric circuit carrying some magnitude of current as well as the resistances is to be remembered. The conceptual relation of Ohm’s law can also be used to resolve the problems related to given conditions. Moreover, the basic concept of Kirchhoff’s Voltage rule is also to be kept in mind.