Question

A direct current of 10 A is superimposed on an alternating current \[I = 40\cos \omega t\](A) flowing through a wire. The effective value of alternating current will be
A. \[10\sqrt 2 \,{\text{A}}\]
B. \[20\sqrt 2 \,{\text{A}}\]
C. \[20\sqrt 3 \,{\text{A}}\]
D. \[30\,{\text{A}}\]

Answer 1

Hint: Determine the rms value of DC current and AC current. Use the formula for the rms value of the current and substitute the values of the AC rms current and DC rms current. For AC current \[I = {I_0}\cos \omega t\]or \[I = {I_0}\sin \omega t\], the rms value is \[\dfrac{{{I_0}}}{{\sqrt 2 }}\].

Formula used:
Effective current, \[I = \sqrt {I_{rms,DC}^2 + I_{rms,AC}^2} \]
Here, \[{I_{rms,DC}}\] is the rms value of DC current and \[{I_{rms,AC}}\] is the rms value of AC current.

Complete step by step answer:
We have given the values of DC current and AC current. The DC current is \[{I_{DC}} = 10\,{\text{A}}\] and AC current is \[{I_{AC}} = 40\cos \omega t\]. Let’s find out the rms value of AC current. The given AC current is, \[{I_{AC}} = 40\cos \omega t\]. Therefore, the rms value of AC current will be,
\[{I_{rms,AC}} = \dfrac{{40}}{{\sqrt 2 }}\]
When the AC current and DC current are superimposed, the resulting rms current is given as,
\[I = \sqrt {I_{rms,DC}^2 + I_{rms,AC}^2} \]

Substituting \[{I_{rms,DC}} = 10\,{\text{A}}\] and \[{I_{rms,AC}} = \dfrac{{40}}{{\sqrt 2 }}\] in the above equation, we get,
\[I = \sqrt {{{\left( {10} \right)}^2} + {{\left( {\dfrac{{40}}{{\sqrt 2 }}} \right)}^2}} \]
\[ \Rightarrow I = \sqrt {100 + \dfrac{{1600}}{2}} \]
\[ \Rightarrow I = \sqrt {100 + 800} \]
\[ \Rightarrow I = \sqrt {900} \]
\[ \therefore I = 30\,{\text{A}}\]
Therefore, the net effective current is 30 A.

So, the correct answer is option D.

Note:The net current is given as the sum of AC current and DC current. The effective value of the current when the AC current and DC current are superimposed is actually given as, \[{I_{eff}} = {\left[ {\dfrac{1}{T}\int\limits_0^T {{I^2}} dt} \right]^{1/2}}\]. Students can also solve these questions by substituting the net current in this equation but it involves a lot of integration. We have discussed the simplest way to solve such a question in the solution part.