A full wave bridge diode rectifier uses a load resistance of 1200 Ω. The internal resistance of each diode is 9 Ω. Find efficiency.

Steps to Solve

1. Understand the given values

Identify the values provided:

• Load resistance, ( R_L = 1200 , \Omega )
• Internal resistance of each diode, ( R_d = 9 , \Omega )

For a full-wave bridge rectifier, there are two diodes in series during conduction. Thus, the total internal resistance is: $$ R_{total_diodes} = 2 \times R_d = 2 \times 9 , \Omega = 18 , \Omega $$

2. Calculate total resistance

Now add the load resistance to the total diode resistance: $$ R_{total} = R_L + R_{total_diodes} $$
Substituting the known values: $$ R_{total} = 1200 , \Omega + 18 , \Omega = 1218 , \Omega $$

3. Calculate the efficiency

The efficiency ( \eta ) of the full-wave rectifier is given by the formula: $$ \eta = \frac{R_L}{R_{total}} \times 100 % $$
Substituting the values: $$ \eta = \frac{1200 , \Omega}{1218 , \Omega} \times 100 % $$

4. Perform the calculation

Calculating the efficiency: $$ \eta \approx \frac{1200}{1218} \times 100 \approx 98.51 % $$