Question

Calculate ripple factor in the following at \(50\ \text{Hz}\):
A. \(\pi\) section filter having \(C_1=200\ \mu F,\ C_2=100\ \mu F,\ L_1=50\ mH,\ R_L=100\ \Omega\),
B. \(\pi\) section filter having \(C_1=200\ \mu F,\ C_2=200\ \mu F,\ L_1=50\ mH,\ R_L=100\ \Omega\),
C. \(\pi\) section filter having \(C_1=200\ \mu F,\ C_2=200\ \mu F,\ L_1=100\ mH,\ R_L=100\ \Omega\),
D. \(\pi\) section filter having \(C_1=100\ \mu F,\ C_2=400\ \mu F,\ L_1=100\ mH,\ R_L=200\ \Omega\). Choose the correct answer from the options given below:

• A. A \(>\) B \(>\) C \(>\) D
• B. D \(>\) C \(>\) B \(>\) A
• C. A \(<\) B \(>\) C \(>\) D
• D. C \(>\) D \(>\) B \(>\) A

Hint:

In filter circuits, larger capacitance, inductance and load resistance generally reduce ripple factor.

Answer 1

The Correct Option is A

Concept:
In a \(\pi\)-section filter, ripple factor decreases when capacitance, inductance and load resistance increase.

Step 1: Understand proportionality.
For comparison, ripple factor is inversely related to the product: \[ C_1C_2L_1R_L \] So, \[ \text{Higher } C_1C_2L_1R_L \Rightarrow \text{Lower ripple factor} \]

Step 2: Compare case A. \[ A\propto 200\times100\times50\times100 \]

Step 3: Compare case B. \[ B\propto 200\times200\times50\times100 \] This product is greater than A, so ripple in B is less than A. \[ A>B \]

Step 4: Compare case C. \[ C\propto 200\times200\times100\times100 \] This product is greater than B, so ripple in C is less than B. \[ B>C \]

Step 5: Compare case D. \[ D\propto 100\times400\times100\times200 \] This product is greatest among the given cases, so ripple in D is least. \[ C>D \] Thus, decreasing order of ripple factor is: \[ A>B>C>D \] \[ \therefore \text{Correct Answer is (A)} \]