Question

A \[100\Omega \] resistance and a capacitor of \[100\Omega \] reactance are connected in series across a \[220V\]source. When the capacitor is 50% charged, the peak value of the displacement current is.

Answer 1

Hint: We will use the method of electrical impedance, which is the resultant sum of the resistance and inductance reactance square. The inductive reactance is at y-axis and resistance is at x-axis when the electrical impedance moves at a certain phase angle, formula is shown as:
\[Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\]
For the current, we use the formula as:
\[I=\dfrac{V}{Z}\]
Where Z is the impedance, R is the resistance value, X is the reactance value, V is the voltage passed through in the series.

Complete step by step answer:
Now to solve the impedance we first place the value of the reactance of the capacitor as \[100\Omega \] and the value of the resistance as \[100\Omega \].
After this we find the impedance like we use the Pythagoras theorem where we take the value of X or reactance as y-axis height and Resistance R as x-axis base length. Placing the values of R and X as \[100\Omega \] in the formula we get the resultant impedance as:
\[Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\]
\[\Rightarrow Z=\sqrt{{{100}^{2}}+{{100}^{2}}}\]
\[\Rightarrow Z=100\sqrt{2}\Omega \]
Now after getting the resultant impedance we find the value of the peak displacement current, unlike any other formula, the value of the current is found by dividing the voltage by the resistance, here the resistance is the net impedance and the voltage is the voltage connected in series as 220V. Placing the value in the formula, we get:
\[I=\dfrac{V}{Z}\]
\[\Rightarrow \dfrac{220}{100\sqrt{2}}\]
\[\therefore 2.2A\]
Therefore, the value of the current passing when the capacitor is \[50%\] charged \[2.2A\].

Note:
We don't use the 50% charged capacitor condition as the voltage will remain constant and doesn't depend upon the charge percentage and the same goes for impedance as well as it will remain the same when the charge is at 100% or at 1%.