Question

A filament bulb ($500\,W,\,100\,V$) is used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
A. $13\Omega $
B. $230\Omega $
C. $46\Omega $
D. $26\Omega $

Answer 1

Hint: Power consumed by an electric appliance is equal to $VI$ where V is the potential difference and I is the current. For resistances connected in series, the sum of potential differences across their ends is equal to the net supply.

Complete step by step answer:
Power consumed by the bulb, $P = V$I
 $\Rightarrow I = \dfrac{P}{V}$
It is given that, Power of the bulb, P = 500 W
Voltage consumed by the bulb, V = 100V
Putting these values in the above relation, we get $I = \dfrac{{500}}{{100}} = 5\,A$
As the resistance R is connected in series with the bulb, so the potential difference across
R = 230 V – 100 V = 130 V
From Ohm’s law we have \[V = IR\]
\[ \Rightarrow IR = 130V \Rightarrow R = \dfrac{{130}}{5} = 26\Omega \]
Hence, the value of R is \[26\Omega. \]

So, the correct answer is “Option D”.

Note:
All electrical appliances such as electric bulbs, electric iron, geysers, toaster, etc are rated in terms of voltage and electric power (wattage). The voltage of an electric appliance is the potential difference that can be safely applied across its input terminals. The electric power of an electric appliance is the rate at which it consumes electrical energy under the rated voltage. For example, a bulb rated at 500 W and 100 V consumes 500 joules per second when the applied voltage is 100 V. On the basis of the rated voltage and the electric power of any electrical appliance, we can calculate its resistance and the current drawn by it.
When two resistances are connected in series, then the sum of potential differences across their ends is equal to the net supply.