Question

$ 10kJ\;{s^{ - 1}} $ of heat is produced when a DC flows through a $ 100\Omega $ resistor. Which of the following ac produces nearly the same heat per sec through the same resistor?
A. $ I = 15\sin \omega t $
B. $ I = 45\sin \omega t $
C. $ I = 30\sin \omega t $
D. $ I = 10\sin \omega t $

Answer 1

Hint: To solve this question, we must have a concept of power decapitated from a resistor and related formula. Here firstly we will with the help of a formula we will find the current drawn and then we will compare with all the options that are giving the same amount of current and from there we will get the required solution.
 $ p = {i^2} \times R $
Where,
 $ p $ is the power
 $ i $ is the current and
 $ R $ is the resistance.

Complete Step By Step Answer:
According to the question it is given that,
 $ p = 10kJ{\text{ }}{s^{ - 1}} $ with the resistor,
 $ R = 100\Omega $
And we know that,
 $ p = {i^2} \times R $
From here we can find the current drawn,
 $
  {i^2} \times 100 = 10000 \\
   \Rightarrow {i^2} = 100 \\
   \Rightarrow i = 10A \\
  $
So, we can get the same RMS value or near about the same as of the current flown from $ 100\Omega $ resistor.
So, now one by one the RMS value of,
 $ 15\sin \omega t = 15\sqrt 2 = 10.6A $
Similarly for second option,
 $ 45\sin \omega t = 45\sqrt 2 = 31.8A $
Again, for third option:
 $ 30\sin \omega t = 30\sqrt 2 = 21.2A $
And the last one,
 $ 10\sin \omega t = 10\sqrt 2 = 7.07A $
So, from the above solution we can conclude that the value of current in the first option is the same or nearly the same to $ 10kJ\;{s^{ - 1}} $ from the $ 100\Omega $ resistor.
Hence, the correct option is A.

Note:
Students usually make mistakes in units. Don’t forget to check the unit as we will use the standard unit in watt. The effective or DC value equivalent for an AC waveform is the rms, or "root mean squared" value. Also, rather than a battery sign for DC, the schematic symbol for defining an AC voltage source is a "wavy" line, as illustrated below.