Question

A direct current of $5A$ is superimposed on an alternating current $I = 10\sin \omega t$ flowing through a wire. The effective value of the resulting current will be:
$(A)\dfrac{{15}}{2}A$
$(B)5\sqrt 3 A$
$(C)5\sqrt 5 A$
$(D)15A$

Answer 1

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:
${I_{rms}} = \sqrt {{{\left( {{I_{DC}}} \right)}^2} + {{\left( {{I_{AC}}} \right)}^2}}$

Complete step-by-step solution:
From given, we have,
The direct current, ${I_{DC}} = 5A$
The alternating current, $I = 10\sin \omega t$
${I_{AC}} = \dfrac{{10}}{{\sqrt 2 }}A$
The effective value of the resulting current is,
${I_{rms}} = \sqrt {{{\left( {{I_{DC}}} \right)}^2} + {{\left( {{I_{AC}}} \right)}^2}}$
$\Rightarrow {I_{eff}} = \sqrt {{{\left( 5 \right)}^2} + {{\left( {\dfrac{{10}}{{\sqrt 2 }}} \right)}^2}}$
$\Rightarrow {I_{eff}} = \sqrt {25 + \dfrac{{100}}{2}}$
$\Rightarrow {I_{eff}} = \sqrt {25 + 50} = \sqrt {75} = \sqrt {3 \times 25}$
$\Rightarrow {I_{eff}} = 3\sqrt 5 A$
A direct current of $5A$ is superimposed on an alternating current $I = 10\sin \omega t$ flowing through a wire. The effective value of the resulting current will be${I_{eff}} = 3\sqrt 5 A$.

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.