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

Steps to Solve

1. Understand the components
   The full-wave bridge rectifier consists of four diodes in configuration. The equivalent resistance of the two conducting diodes (since two diodes conduct at a time) is given by:
   $$ R_d = 2 \times 9 , \Omega = 18 , \Omega $$

2. Determine total resistance
   The total resistance in the circuit can be found by adding the load resistance ($R_L$) and the equivalent resistance of the diodes ($R_d$):
   $$ R_{total} = R_L + R_d = 1200 , \Omega + 18 , \Omega = 1218 , \Omega $$

3. Calculate the output voltage
   Assuming the input voltage ($V_s$) is the peak voltage of the AC supply, the output voltage across the load resistance ($V_{out}$) is then approximately:
   $$ V_{out} = V_s - 2V_f $$
   where $V_f$ is the forward voltage drop across each diode. Assuming $V_f \approx 0.7 , V$, the equation becomes:
   $$ V_{out} = V_s - 2(0.7) = V_s - 1.4 , V $$

4. Calculate load current
   The load current ($I_L$) can be calculated using Ohm's law:
   $$ I_L = \frac{V_{out}}{R_L} = \frac{V_s - 1.4}{1200} $$

5. Determine output power ($P_{out}$)
   The output power can be expressed as:
   $$ P_{out} = I_L^2 \cdot R_L $$
   Substituting $I_L$ we have:
   $$ P_{out} = \left(\frac{V_s - 1.4}{1200}\right)^2 \cdot 1200 $$

6. Determine input power ($P_{in}$)
   The power dissipated in the internal diode resistance can be calculated as:
   $$ P_{in} = I_L^2 \cdot R_{total} = \left(\frac{V_s - 1.4}{1200}\right)^2 \cdot 1218 $$

7. Calculate efficiency
   Finally, the efficiency ($\eta$) can be calculated as:
   $$ \eta = \frac{P_{out}}{P_{in}} \times 100% $$

Since $P_{out}$ and $P_{in}$ have a common term, we can simplify efficiency calculation into:
$$ \eta = \frac{R_L}{R_{total}} \times 100% $$
Substituting the values gives:
$$ \eta = \frac{1200}{1218} \times 100% $$