VS 7.07 Vrms PUT R1 1K VGG Max 9V Question 1 Compute the firing angle, if VGG is set to 50%

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VS 7.07 Vrms PUT R1 1K VGG Max 9V Question 1 Compute the firing angle, if VGG is set to 50%. Do not include the units, the final answer must be in 2 decimal places. Question 2 Compute the firing angle, if VGG is set to 80%. Do not include the units, the final answer must be in 2 decimal places.

The Correct Answer and Explanation is:

Let’s solve both questions step by step.

Given:

  • Vrms=7.07V_{rms} = 7.07Vrms​=7.07 V
  • R1=1R_1 = 1R1​=1 kΩ
  • VGGV_{GG}VGG​ (control voltage) can be set to 50% and 80%.
  • The maximum VGGV_{GG}VGG​ is 9V.

We are asked to compute the firing angle (α\alphaα) in two scenarios.

Formula for the Firing Angle:

The firing angle α\alphaα is related to the control voltage VGGV_{GG}VGG​ in a phase-controlled rectifier circuit. The relationship can be described as:VGG=Vmax×(1−cos⁡α)V_{GG} = V_{max} \times (1 – \cos \alpha)VGG​=Vmax​×(1−cosα)

Where:

  • VmaxV_{max}Vmax​ is the peak value of the voltage (i.e., Vmax=Vrms×2V_{max} = V_{rms} \times \sqrt{2}Vmax​=Vrms​×2​).

Step 1: Find VmaxV_{max}Vmax​

Since Vrms=7.07V_{rms} = 7.07Vrms​=7.07 V,Vmax=7.07×2=7.07×1.414=10 VV_{max} = 7.07 \times \sqrt{2} = 7.07 \times 1.414 = 10 \, \text{V}Vmax​=7.07×2​=7.07×1.414=10V

Thus, Vmax=10V_{max} = 10Vmax​=10 V.

For Question 1: When VGG=50%V_{GG} = 50\%VGG​=50% of 9V

This means VGG=0.5×9=4.5V_{GG} = 0.5 \times 9 = 4.5VGG​=0.5×9=4.5 V.

Now, substitute this into the formula:VGG=Vmax×(1−cos⁡α)V_{GG} = V_{max} \times (1 – \cos \alpha)VGG​=Vmax​×(1−cosα)4.5=10×(1−cos⁡α)4.5 = 10 \times (1 – \cos \alpha)4.5=10×(1−cosα)

Solving for cos⁡α\cos \alphacosα:1−cos⁡α=4.510=0.451 – \cos \alpha = \frac{4.5}{10} = 0.451−cosα=104.5​=0.45cos⁡α=1−0.45=0.55\cos \alpha = 1 – 0.45 = 0.55cosα=1−0.45=0.55

Now, calculate α\alphaα:α=cos⁡−1(0.55)≈56.25∘\alpha = \cos^{-1}(0.55) \approx 56.25^\circα=cos−1(0.55)≈56.25∘

So, the firing angle is 56.25 degrees.

For Question 2: When VGG=80%V_{GG} = 80\%VGG​=80% of 9V

This means VGG=0.8×9=7.2V_{GG} = 0.8 \times 9 = 7.2VGG​=0.8×9=7.2 V.

Substitute into the formula:VGG=Vmax×(1−cos⁡α)V_{GG} = V_{max} \times (1 – \cos \alpha)VGG​=Vmax​×(1−cosα)7.2=10×(1−cos⁡α)7.2 = 10 \times (1 – \cos \alpha)7.2=10×(1−cosα)

Solving for cos⁡α\cos \alphacosα:1−cos⁡α=7.210=0.721 – \cos \alpha = \frac{7.2}{10} = 0.721−cosα=107.2​=0.72cos⁡α=1−0.72=0.28\cos \alpha = 1 – 0.72 = 0.28cosα=1−0.72=0.28

Now, calculate α\alphaα:α=cos⁡−1(0.28)≈73.74∘\alpha = \cos^{-1}(0.28) \approx 73.74^\circα=cos−1(0.28)≈73.74∘

So, the firing angle is 73.74 degrees.

Final Answers:

  • For Question 1: The firing angle is 56.25 degrees.
  • For Question 2: The firing angle is 73.74 degrees.

These calculations assume ideal conditions where there is no significant loss or variation in the system.

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