Question

a. Evaluate $V_C$at emf frequencies $1,3,10,30,100KHz$

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

Short answer

Part(a)$V_C$at emf frequencies $1,3,10,30,100KHz$are $9.94,9.57,7.05,3,15,0.99V$respectively.

Part(b) Required figure is given below.

Step-by-step solution

Part(a) Step 1 : Given information

We are given various frequencies $1,3,10,30,100KHz$. with Capacitance of $1.0\mu F$, Resistance of $16\Omega$, Emf of $10V\left(\cos \left(wt\right)\right)$

We have to find $V_c$

Part(a) Step 2 : Simplify

As we know,

$V_c=I{X}_c$

$V_C=$voltage across capacitor

$f=$frequency

$R=$resistance

$C=$capacitance

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

Where, the capacitive reactance is

$X_c=\frac{1}{wC}=\frac{1}{2\pi fC}$

Now,

the peak current is

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

$V_C=\frac{{\in }_{\circ }}{\sqrt{1+{\left(2\pi fRC\right)}^2}}$

a.

If $f=1KHz$

$V_c=\frac{10}{\sqrt{1+{\left(2\pi \left(1\right)\left(16\right)\left(1\times {10}^{-3}\right)\right)}^2}}=9.94V$

b.

If $f=3KHz$

$V_C=\frac{10}{\sqrt{1+{\left(2\pi \left(3\right)\left(16\right)\left(1\times {10}^{-3}\right)\right)}^2}}=9.57V$

c.

If $f=10KHz$

$V_C=\frac{10}{\sqrt{1+{\left(2\pi \left(10\right)\left(16\right)\left(1\times {10}^{-3}\right)\right)}^2}}=7.05V$

d.

If $f=30KHz$

$V_C=\frac{10}{\sqrt{1+{\left(2\pi \left(30\right)\left(16\right)\left(1\times {10}^{-3}\right)\right)}^2}}=3.15V$

e.

If $f=100KHz$

$V_C=\frac{10}{\sqrt{1+{\left(2\pi \left(100\right)\left(16\right)\left(1\times {10}^{-3}\right)\right)}^2}}=0.99V$

Part(b) Step 1 : Given information

We have to draw graph between $V_C$vs Frequencies

Part(b) Step 2 : Simplify

Graph