Question

A low-pass $RC$ filter has a crossover frequency $f_c=200$Hz.What is $f_c$if:

a. The resistance $R$is doubled?

b. The capacitance $C$is doubled?

c. The peak emf $E_0$is doubled?

Short answer

Part(a) $f_c$will be $100$Hz

Part(b) $f_c$will be $100$Hz

Part(c) $f_c$will be $200$Hz

Step-by-step solution

Part(a) Step 1 : Given information

We have been given that

A low-pass $RC$ filter has a crossover frequency $f_c=200$Hz

We have to find $f_c$if resistance $R$is doubled.

Part(a) Step 2 : Simplify

As we know in Low Pass$RC$ filter ,

$f_c=\frac{1}{2\pi RC}$

where,

$R=$Resistance

$C=$Capacitance

Now, according to question:

$f_c=\frac{1}{2\pi RC}=200$Hz

if resistance is doubled

$R$becomes $2R$

${\left(f_c\right)}^{\text{'}}=\frac{1}{2\pi \left(2R\right)C}=\frac{f_c}{2}=\frac{200}{2}=100$Hz

so $f_c$becomes $100$Hz if resistance $R$is doubled.

Part(b) Step 1 : Given information

We have been given that

A low-pass $RC$ filter has a crossover frequency $f_c$$=200$Hz.

We have to find $f_c$if capacitance $C$is doubled.

Part(b) Step 2 : Simplify

As we know in Low Pass $RC$filter ,

$f_c=\frac{1}{2\pi RC}$

where,

$R=$resistance

$C=$capacitance

Now, according to question:

$f_c=\frac{1}{2\pi RC}=200$Hz

If resistance is doubled

$C$becomes $2C$

${\left(f_c\right)}^{\text{'}}=\frac{1}{2\pi R\left(2C\right)}=\frac{f_c}{2}=\frac{200}{2}=100$Hz

so $f_c$becomes $100$Hz if capacitance $C$is doubled.

Part(c) Step 1 : Given information

We have been given that

A low-pass $RC$filter has a crossover frequency $f_c$$=200$Hz

We have to find $f_c$if peak emf $E_0$is doubled.

Part(c) Step 2 : Simplify

As we know in Low Pass $RC$filter ,

$f_c=\frac{1}{2\pi RC}$

where,

$R=$Resistance

$C=$Capacitance

Now, according to question

If peak emf $E_0$is doubled there will be no change.