What are$V_{\mathrm{R}}$and $V_{\mathrm{C}}$if the emf frequency in FIGURE is $10\mathrm{kHz}$?

The ampere is the SI unit for current, and it quantifies the passage of electric charge over a surface at one coulomb per second. The unit is coulomb/Sec (C/s) or amp because the charge is evaluated in coulombs and the duration is measured in seconds.

Assumed values:

$f=10\mathrm{kHz}\to 1\times {10}^4\mathrm{Hz}$(hertz)

$R=150\Omega$(ohm)

$C=80\times {10}^{-9}\mathrm{F}$(farad)

${\epsilon }_o=10\cos wt$

localid="1650216237862" $w=2\pi f$

$=2\pi \times {10}^4\mathrm{rad}/\mathrm{s}$

The peak current is given by ,

$I=\frac{{\epsilon }_0}{\sqrt{R^2+X_C^2}}$

From there, we come across

$I=\frac{{\epsilon }_0}{\sqrt{R^2+X_C^2}}$

localid="1650213010427" $I=\frac{10V}{\sqrt{{150}^2\Omega +{\left(2\pi \left({10}^4\right)\times 80\times {10}^{-9}\right)}^{-2}F}}$

$I=0.04A$

We then calculate the captivate reactance $V_c$Where$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

$V_C=I{X}_C$

In the equation, substitute values

localid="1650212969080" $V_C=\frac{0.04A}{2\pi \left({10}^4\right){\left(80\times {10}^{-2}\right)}^{}F}$

$V_C=8V$

The Voltage are $V_R=IR$:

$V_R=IR$

Fill in the blanks in the equation,

$V_R=0.04\mathrm{A}\times 150\Omega$

$V_R=6\mathrm{V}$