This problem involves a single-phase rectifier bridge with thyristor control. Let's break down the solution into parts:

Given dataThis problem involves a single-phase rectifier bridge with thyristor control. Let's break down the solution into parts:

Given

• Peak voltage $E_m$ = 340 V
• AC frequency $f$ = 50 Hz
• Load resistance $R$ = 5 \Omega
• DC output voltage $V_{DC}$ = 100 V
• Power consumed by the DC source $P$ = 1 kW (1000 W)
• Current ripple constraint: Peak-to-peak current ripple should not exceed 2A

We'll work through the problem step by step.

a) Find the firing angle $\alpha$

To find the firing angle $\alpha$, we need to calculate the voltage and power relationships in the circuit.

1. Output voltage for a controlled rectifier:

   The average output voltage for a fully controlled rectifier bridge is given by: $V_{DC} = \frac{2E_m}{\pi} \cdot \cos(\alpha)$ Where $\alpha$ is the firing angle, and $E_m$ is the peak voltage of the AC supply.

2. Power consumption:

   The power consumed by the DC voltage source is given by: $P = V_{DC} \cdot I_{DC}$ Rearranging for the current $I_{DC}$: $I_{DC} = \frac{P}{V_{DC}} = \frac{1000 \, \text{W}}{100 \, \text{V}} = 10 \, \text{A}$ So, the average current $I_{DC}$ is 10 A.

3. Determine the DC voltage:

   Using the equation for $V_{DC}$ in terms of $\alpha$: $100 = \frac{2 \times 340}{\pi} \cdot \cos(\alpha)$ Solving this equation for $\alpha$: $\cos(\alpha) = \frac{100 \pi}{2 \times 340}$

Let's compute $\alpha$.The firing angle $\alpha$ is approximately $62.48^\circ$.

b) Size of the coil (inductance $L$) to limit peak-to-peak current ripple

To ensure that the peak-to-peak current ripple does not exceed 2 A, we need to calculate the required inductance $L$.

1. Ripple current equation: The current ripple $\Delta I$ in a rectifier circuit is given by: $\Delta I = \frac{V_{peak} \cdot (T/2)}{L}$ Where:

   • $V_{peak}$ is the peak AC voltage,
   • $T$ is the period of the AC signal (in this case, $T = 1/f$),
   • $L$ is the inductance.

   The peak-to-peak current ripple is specified as 2 A. Hence, $\Delta I = 2 \, \text{A}$.

2. Determine the AC voltage across the inductor: In a rectified circuit, the voltage across the inductor is approximately $V_{peak} = E_m - V_{DC}$.

3. Substitute values: $L = \frac{(E_m - V_{DC}) \cdot (T/2)}{\Delta I}$ Where $T = \frac{1}{f} = \frac{1}{50} \, \text{s}$.

Let's calculate the required inductance $L$.The required inductance $L$ to ensure that the peak-to-peak current ripple does not exceed 2 A is $1.2 \, \text{H}$.

Summary of Results:

a) The firing angle $\alpha$ is approximately $62.48^\circ$.
b) The inductance $L$ should be $1.2 \, \text{H}$ to limit the peak-to-peak current ripple to 2 A.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How does increasing the inductance affect the current ripple in this circuit?
2. What happens to the firing angle if the DC voltage is increased?
3. How would the power factor be affected by a change in the firing angle?
4. Can we achieve better ripple control with other types of filters besides the inductor?
5. How does the frequency of the AC source impact the ripple current?

Tip: Increasing inductance in the output smooths the current more, reducing ripple but increasing response time.