A signal generator supplies a sine wave of $200V$, $5 kHz$ to the circuit shown in the figure. Thus choose the wrong statement

$\begin{array}{rcl}{I}_R& =& \frac{V_{rms}}{R}=\frac{200}{100}=2 \mathrm{A}\\ X_C& =& \frac{1}{2\pi fC}=\frac{1}{\left(2\pi \right)(5\times {10}^3)({\displaystyle \frac{1}{\pi }}\times {10}^{-6})}\\ & =& 100\Omega \\ \therefore I_C& =& \frac{V_{rms}}{X_C}=\frac{200}{100}=2 \mathrm{A}\\ & & I_C {\text{is 90° ahead of the applied voltage and I}}_{\mathrm{R}}\text{ is in }\\ & & \text{phase with the applied voltage. Hence, there is a }\\ & & \text{phase difference of }\mathrm{90°} \mathrm{between} {\mathrm{I}}_{\mathrm{R} }\mathrm{and} {\mathrm{I}}_{\mathrm{C} }\mathrm{too}.\\ \therefore I_{\text{total} }& =& \sqrt{I_R^2+I_C^2}\\ & =& \sqrt{{\left(2\right)}^2+{\left(2\right)}^2}\\ & =& 2.83 \mathrm{A}\end{array}$