Question

a. Evaluate $V_R$ in figure at emf frequencies$100,300,1000,3000,10000Hz$

b. Graph $V_R$versus frequency. Draw a smooth curve through your five points.

Short answer

Part(a)$V_R$at emf frequencies $100,300,1000,3000,10000$are $1.00,2.90,7.11,9.47,9.97V$ respectively.

Part(b) Required figure is given below.

Step-by-step solution

Part(a) Step 1: Given information

We have given different frequencies

$100,300,1000,3000,10000$and $R=100\Omega ,E_0=10V,C=1.6\times {10}^{-3}F$

We have to find $V_R$

Part(a) Step 2 : Simplify

As we know,

$V_R=$voltage across resistor

$R=$resistance

$C=$capacitance

$f=$frequency

$X_C=$capacitive reactance

${\in }_{\circ }=$emf

$$\begin{gathered} V_R=IR \\ I=\frac{{{\rm E}}_0}{\sqrt{R^2+X_C^2}} \\ {X}_C=\frac{1}{WC}=\frac{1}{2\pi fC} \end{gathered}$$

Now if

a. $$\begin{gathered} f=100Hz \\ {V}_R=\frac{{\in }_{\circ }}{\sqrt{R^2+{\displaystyle \frac{1}{{\left(2\pi fC\right)}^2}}}}=\frac{10}{\sqrt{{100}^2+{\displaystyle \frac{1}{{\left(2\pi \left(100\right)\left(1.6\times {10}^{-3}\right)\right)}^2}}}} \\ {V}_R=1.00V \\ f=300Hz \\ {V}_R=\frac{{\in }_{\circ }}{\sqrt{R^2+{\displaystyle \frac{1}{{\left(2\pi fC\right)}^2}}}}=\frac{10}{\sqrt{{100}^2+{\displaystyle \frac{1}{{\left(2\pi \left(300\right)\left(1.6\times {10}^{-3}\right)\right)}^2}}}} \\ {V}_R=2.90V \\ f=1000Hz \\ {V}_R=\frac{{\in }_{\circ }}{\sqrt{R^2+{\displaystyle \frac{1}{{\left(2\pi fC\right)}^2}}}}=\frac{10}{\sqrt{{100}^2+{\displaystyle \frac{1}{{\left(2\pi \left(1000\right)\left(1.6\times {10}^{-3}\right)\right)}^2}}}} \\ {V}_R=7.11V \\ f=3000Hz \\ {V}_R=\frac{{\in }_{\circ }}{\sqrt{R^2+{\displaystyle \frac{1}{{\left(2\pi fC\right)}^2}}}}=\frac{10}{\sqrt{{100}^2+{\displaystyle \frac{1}{{\left(2\pi \left(3000\right)\left(1.6\times {10}^{-3}\right)\right)}^2}}}} \\ {V}_R=9.47V \\ f=10000Hz \\ {V}_R=\frac{{\in }_{\circ }}{\sqrt{R^2+{\displaystyle \frac{1}{{\left(2\pi fC\right)}^2}}}}=\frac{10}{\sqrt{{100}^2+{\displaystyle \frac{1}{{\left(2\pi \left(10000\right)\left(1.6\times {10}^{-3}\right)\right)}^2}}}} \\ {V}_R=9.97V \end{gathered}$$

Part(b) Step 1: Given information

We have to draw graph $V_R$ vs frequencies

Part(b) Step 2 : Simplify

Graph: