Question

The instantaneous voltage through a device of impedance $20\Omega $ is, $e=80\sin 100\pi t$. The effective value of the current is
A. 3A
B. 2.828A
C. 1.732A
D. 4A

Answer 1

Hint: The effective value of an alternating current is the RMS value of the current. RMS is an abbreviation used for root mean square value of the current or voltage. RMS value of AC current is equivalent to the DC constant value of the varying current or voltage.

Complete step by step answer:
According to given data,
Impedance (Z) = 20Ω
$e=80\sin 100\pi t.........(a)$
Formula to be considered:
$\begin{align}
  & e={{e}_{m}}\sin \omega t...........(1) \\
 & {{i}_{rms}}=\dfrac{{{i}_{m}}}{\sqrt{2}}.............(2) \\
\end{align}$
The AC current is a type of current that changes its direction periodically (from positive half to the negative half). Due to the periodic behavior the average of current and voltage over a one complete cycle is zero. To overcome this the voltage and current are measured only for the positive half cycle of the wave. The general equation for the AC current and voltage are:
$\begin{align}
  & e={{e}_{m}}\sin \omega t \\
 & i={{i}_{m}}\sin \omega t \\
\end{align}$
Where em and im are the peak voltage and current respectively.
By comparing equations (a) and (1), we can say that the peak value of voltage for the given condition would be,
${{e}_{m}}=80$
Using the estimated value of peak voltage. The peak current of the given circuit will be:
$\begin{align}
  & {{i}_{m}}=\dfrac{{{e}_{m}}}{Z} \\
 & \Rightarrow \dfrac{80}{20} \\
 & \Rightarrow 4A \\
\end{align}$
To know the effective voltage and current of the circuit RMS (root mean square) value of the current and voltage is calculated. Which is given by:
$\begin{align}
  & {{e}_{rms}}=\dfrac{{{e}_{m}}}{\sqrt{2}} \\
 & {{i}_{rms}}=\dfrac{{{i}_{m}}}{\sqrt{2}} \\
\end{align}$
Thus, the effective current across the provided circuit condition will be the RMS value of the peak current.
 From equation (2) and the calculated value of peak current, the RMS value of current will be:
$\begin{align}
  & {{i}_{rms}}=\dfrac{{{i}_{m}}}{\sqrt{2}} \\
 & \Rightarrow \dfrac{4}{\sqrt{2}}=2\sqrt{2} \\
 & \Rightarrow 2.8284A \\
\end{align}$
So the correct option which represents the effective current flowing through the circuit is option B and the value of effective current is 2.828A.

Note:
The peak current and the RMS value of current are two different properties. RMS value of current represents the effective current across the circuit and the peak value defines the maximum current value which can be attained by the circuit.