Sure, let's solve the problem step by step.
### Step 1: Calculate the Peak Load Current
Given:
- Transformer peak voltage ([tex]\(V_{peak}\)[/tex]): 30V
- Load resistance ([tex]\(R_L\)[/tex]): 100 ohms
- Diode forward resistance ([tex]\(R_D\)[/tex]): 10 ohms
First, we need to determine the total resistance in the circuit. This includes both the load resistance and the forward resistance of the diode:
[tex]\[ R_{total} = R_L + R_D \][/tex]
Substituting the given values:
[tex]\[ R_{total} = 100 \Omega + 10 \Omega = 110 \Omega \][/tex]
Using Ohm's Law, the peak load current ([tex]\(I_{peak}\)[/tex]) can now be calculated by:
[tex]\[ I_{peak} = \frac{V_{peak}}{R_{total}} \][/tex]
Substituting the given peak voltage and total resistance:
[tex]\[ I_{peak} = \frac{30V}{110 \Omega} \approx 0.2727 \, A \][/tex]
### Step 2: Calculate the Average Load Current
For a full-wave rectifier, the average load current ([tex]\(I_{avg}\)[/tex]) is given by:
[tex]\[ I_{avg} = \frac{2 \cdot I_{peak}}{\pi} \][/tex]
Substituting the peak current calculated earlier:
[tex]\[ I_{avg} = \frac{2 \cdot 0.2727}{\pi} \approx 0.1736 \, A \][/tex]
### Step 3: Calculate the Average Load Voltage
The average load voltage ([tex]\(V_{avg}\)[/tex]) can be determined using:
[tex]\[ V_{avg} = I_{avg} \cdot R_L \][/tex]
Substituting the average current and load resistance:
[tex]\[ V_{avg} = 0.1736 \, A \cdot 100 \Omega = 17.36 \, V \][/tex]
### Step 4: Calculate the Rectifier Efficiency
First, we calculate the input power:
The transformer input power ([tex]\(P_{in}\)[/tex]) can often be estimated by considering the average power delivered through the load and the resistances. If simplified, for ideal conditions, the input power can be expressed as:
[tex]\[ P_{in} = \frac{V_{peak}^2}{2 \cdot R_{total}} \][/tex]
Substituting the values:
[tex]\[ P_{in} = \frac{30^2}{2 \cdot 110} = \frac{900}{220} \approx 4.09 \, W \][/tex]
Next, calculate the output power delivered to the load:
[tex]\[ P_{out} = V_{avg} \cdot I_{avg} \][/tex]
Substituting the average voltage and current:
[tex]\[ P_{out} = 17.36 \, V \cdot 0.1736 \, A \approx 3.01 \, W \][/tex]
The rectifier efficiency ([tex]\(\eta\)[/tex]) is then given by:
[tex]\[ \eta = \frac{P_{out}}{P_{in}} \times 100 \][/tex]
Substituting the calculated input and output power:
[tex]\[ \eta = \frac{3.01 \, W}{4.09 \, W} \times 100 \approx 73.69\% \][/tex]
### Final Results
So the average load current, average load voltage, and rectifier efficiency are:
1. Average Load Current: [tex]\(0.1736 \, A\)[/tex]
2. Average Load Voltage: [tex]\(17.36 \, V\)[/tex]
3. Rectifier Efficiency: [tex]\(73.69\% \)[/tex]
