Question

A direct current of 5 A is superimposed on an alternating current \[I=10\sin wt\] flowing through a wire. The effective value of the resulting current will be:
\[A.\,({15}/{2}\;)\,A\]
\[B.\,5\sqrt{3}\,A\]
\[C.\,5\sqrt{5}\,A\]
\[D.\,15\,A\]

Answer 1

Hint: The total value of the resulting current will be the sum of DC value and AC value, whereas, the effective value of the resulting current will be its RMS value (the root of the sum of the squares of DC and AC values). There are two methods to solve this problem.

Formula used:
\[\begin{align}
  & {{I}_{total}}={{I}_{DC}}+{{I}_{AC}} \\
 & {{I}_{rms}}=\sqrt{{{({{I}_{DC}})}^{2}}+{{({{I}_{AC}})}^{2}}} \\
 & {{I}_{eff}}={{\left[ \dfrac{\int\limits_{0}^{T}{{{I}^{2}}dt}}{\int\limits_{0}^{T}{dt}} \right]}^{\dfrac{1}{2}}} \\
\end{align}\]

Complete answer:
There are two methods to solve this problem.

Method I:

From given, we have,
The direct current, \[{{I}_{DC}}=5\,A\]
The alternating current, \[{{I}_{AC}}=10\sin wt\]
\[\Rightarrow {{I}_{AC}}=\dfrac{10}{\sqrt{2}}\,\,A\]
The effective value of the resulting current is,
\[\begin{align}
  & {{I}_{eff}}=\sqrt{{{({{I}_{DC}})}^{2}}+{{({{I}_{AC}})}^{2}}} \\
 & {{I}_{eff}}=\sqrt{{{(5)}^{2}}+{{\left( \dfrac{10}{\sqrt{2}} \right)}^{2}}} \\
 & {{I}_{eff}}=\sqrt{25+\left( \dfrac{100}{2} \right)} \\
 & {{I}_{eff}}=5\sqrt{3}\,A \\
\end{align}\]

Method II:

From given, we have,
The direct current, \[{{I}_{DC}}=5\,A\]
The alternating current, \[{{I}_{AC}}=10\sin wt\]
\[\Rightarrow {{I}_{AC}}=\dfrac{10}{\sqrt{2}}\,\,A\]
The effective value of the resulting current is,
The total current in the wire is,
\[{{I}_{total}}={{I}_{DC}}+{{I}_{AC}}\]
Substitute the given values in the above equation.
\[{{I}_{total}}=5+10\sin wt\]
Compute the effective value of the resulting current
\[{{I}_{eff}}={{\left[ \dfrac{\int\limits_{0}^{T}{{{I}^{2}}dt}}{\int\limits_{0}^{T}{dt}} \right]}^{\dfrac{1}{2}}}\]
Substitute the values in the above equation.
\[{{I}_{eff}}={{\left[ \dfrac{1}{T}\int\limits_{0}^{T}{{{(5+10\sin wt)}^{2}}dt} \right]}^{\dfrac{1}{2}}}\]
First, perform the algebraic operation, that is, \[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[{{I}_{eff}}={{\left[ \dfrac{1}{T}\int\limits_{0}^{T}{(25+100\sin wt+100{{\sin }^{2}}wt)dt} \right]}^{\dfrac{1}{2}}}\]
Now, separately perform the integration operation on all the above terms and apply the limits.
\[{{I}_{eff}}={{\left[ 25+0+(100)\dfrac{1}{2} \right]}^{\dfrac{1}{2}}}\]

Further, solve the above equation to obtain the result.
\[{{I}_{eff}}=5\sqrt{3}\,A\]

So, the correct answer is “Option B”.

Note:
The things to be on your figure tips for further information on solving these types of problems are:
The effective current is the same or is equal to the rms current. The rms value is the effective value of the complete waveform, whereas, the peak value is the highest value that a wave will ever reach.
\[{{I}_{rms}}=\dfrac{I{}_{0}}{\sqrt{2}}\], where \[{{I}_{rms}}\] is the rms value and \[I{}_{0}\] is the peak value.