EEE (Spring - 2023)

(1)

(a) Write the condition for resonance for both series and parallel RLC circuits. Prove that for resonance frequency f a = 1 2pi sqrt 1.C f c both series and parallel RLC circuits.

(a) Condition for Resonance:

Series RLC Circuit:
Resonance occurs when the inductive reactance equals the capacitive reactance:

$$X_L = X_C \quad \text{or} \quad \omega L = \frac{1}{\omega C}.$$

Parallel RLC Circuit:
Resonance occurs when the total impedance is purely resistive, and the admittance of inductance and capacitance cancel each other:

$$\frac{1}{X_L} = \frac{1}{X_C} \quad \text{or} \quad \omega L = \frac{1}{\omega C}.$$

Resonance Frequency Proof:
At resonance, for both series and parallel RLC circuits:

$$\omega = \frac{1}{\sqrt{LC}}.$$

Convert angular frequency ($\omega$) to regular frequency ($f$):

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi \sqrt{LC}}.$$

Thus, the resonance frequency for both circuits is:

$$f=\frac{1}{2\pi \sqrt{LC}}\mathrm{.}$$

(a) Condition for Resonance:

Series RLC Circuit:
Resonance occurs when the inductive reactance equals the capacitive reactance:

$$X_L = X_C \quad \text{or} \quad \omega L = \frac{1}{\omega C}.$$

Parallel RLC Circuit:
Resonance occurs when the total impedance is purely resistive, and the admittance of inductance and capacitance cancel each other:

$$\frac{1}{X_L} = \frac{1}{X_C} \quad \text{or} \quad \omega L = \frac{1}{\omega C}.$$

Resonance Frequency Proof:
At resonance, for both series and parallel RLC circuits:

$$\omega = \frac{1}{\sqrt{LC}}.$$

Convert angular frequency ($\omega$) to regular frequency ($f$):

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi \sqrt{LC}}.$$

Thus, the resonance frequency for both circuits is:

$$f=\frac{1}{2\pi \sqrt{LC}}\mathrm{.}$$

(2)

(a) A positive-sequence, balanced A-connected source supplies a balanced A connected load. If the impedance per phase of the load is 18+j12 and I = 22.5<35-A, find Iab and Vab 

 (b) A balanced Y-connected load with a phase resistance of 40 and a reactance of 250 is supplied by a balanced, positive sequence A-connected source with a line voltage of 210 V. Calculate the phase currents. Use Va as reference.

(a) Finding $I_{ab}$ and $V_{ab}$:

Given:

• Load impedance per phase, $Z = 18 + j12$
• Phase current, $I = 22.5 \angle 35^\circ \, \text{A}$

1. Calculate Line Current ($I_{ab}$):
   In $\Delta$-connected systems, the line current ($I_{ab}$) is related to the phase current ($I$) as:

$$I_{ab} = \sqrt{3} \times I \, \angle (35^\circ + 30^\circ)$$ $$I_{ab} = \sqrt{3} \times 22.5 \, \angle 65^\circ$$ $$I_{ab} = 38.96 \, \angle 65^\circ \, \text{A}$$

2. Calculate Line Voltage ($V_{ab}$):
   In $\Delta$-connected systems, line voltage equals phase voltage. Using $V = Z \times I$:

$$V = (18 + j12) \times 22.5 \angle 35^\circ$$ $$V = 22.5 \times 21.63 \angle (35^\circ + \tan^{-1} \frac{12}{18})$$ $$V = 486.75 \, \angle 78.69^\circ \, \text{V}$$

Thus, $V_{ab} = 486.75 \, \angle 78.69^\circ \, \text{V}$.

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(b) Finding Phase Currents in Y-Connection:

Given:

• Line voltage, $V_L = 210 \, \text{V}$
• Load impedance per phase, $Z = 40 + j250$
• $\Delta$-connected source with a balanced Y-connected load.

1. Phase Voltage ($V_\phi$):
   In Y-connection, the phase voltage is:

$$V_\phi = \frac{V_L}{\sqrt{3}} = \frac{210}{\sqrt{3}} = 121.24 \, \text{V}.$$

2. Phase Current ($I_\phi$):
   Using Ohm’s law:

$$I_\phi = \frac{V_\phi}{Z} = \frac{121.24}{\sqrt{40^2 + 250^2}} \, \angle (-\tan^{-1} \frac{250}{40})$$ $$I_\phi = \frac{121.24}{253.2} \, \angle -80.9^\circ = 0.479 \, \angle -80.9^\circ \, \text{A}.$$

Thus, phase current:

$$I_{\varphi }=0.479\mathrm{\angle }-80.9^{\circ }\text{A}\mathrm{.}$$

(3) (OR)

b) The two-wattmeter method produces wattmeter readings P * 1 = 1560W and P * 2 = 2100 W when connected to a delta-connected load. The line voltage is 220v, 

i) Draw the circuit connection, 

(ii) the per-phase average power, 

(iii) the per-phase reactive power, 

(iv) the power factor, and 

(v) the phase impedance.

(b)

(i) Circuit Diagram:

A simple circuit showing a delta-connected load with two wattmeters connected in two lines of the three-phase supply.

(ii) Per-phase average power:

Total power = $P_1 + P_2 = 1560 + 2100 = 3660 \, \text{W}$
Per-phase average power = $\frac{\text{Total power}}{3} = \frac{3660}{3} = 1220 \, \text{W}$

(iii) Per-phase reactive power:

Reactive power, $Q = \sqrt{3} \times (P_1 - P_2)$
$Q = 1.732 \times (2100 - 1560) = 936.2 \, \text{VAR}$
Per-phase reactive power = $\frac{Q}{3} = \frac{936.2}{3} \approx 312.1 \, \text{VAR}$

(iv) Power Factor (PF):

$\text{PF} = \frac{\text{Total Power}}{\sqrt{3} \times \text{Line Voltage} \times \text{Line Current}}$
Using formula $\cos \phi = \frac{P}{\sqrt{P^2 + Q^2}}$:
$\text{PF} = \frac{3660}{\sqrt{3660^2 + 936.2^2}} \approx 0.97$

(v) Phase Impedance:

Using $Z = \frac{V_{ph}}{I_{ph}}$:
Phase voltage $V_{ph} = \frac{V_{L}}{\sqrt{3}} = \frac{220}{1.732} \approx 127 \, \text{V}$
Phase current $I_{ph} = \frac{P_{ph}}{V_{ph}} = \frac{1220}{127} \approx 9.61 \, \text{A}$
$Z=\frac{127}{9.61}\approx 13.2\mathrm{\Omega }$

(4)

a) Differentiate between a normal transformer and an auto transformer. Draw the arrangement for the step-up and step-down auto transformers.

(a)

Difference between Normal Transformer and Auto Transformer:

1. Construction:

   • Normal Transformer: Has two separate windings (primary and secondary).
   • Auto Transformer: Has one continuous winding with a part used as primary and secondary.

2. Size and Cost:

   • Normal Transformer: Larger and more expensive.
   • Auto Transformer: Smaller and cheaper.

3. Efficiency:

   • Normal Transformer: Slightly less efficient due to higher copper losses.
   • Auto Transformer: More efficient due to reduced losses.

4. Voltage Range:

   • Normal Transformer: Used for wide voltage differences.
   • Auto Transformer: Suitable for small voltage differences.

5. Isolation:

   • Normal Transformer: Provides electrical isolation.
   • Auto Transformer: Does not provide isolation.

Arrangements:

Step-Up Auto Transformer:

• Low voltage is applied to a part of the winding; the output is taken from the entire winding.

Step-Down Auto Transformer:

• High voltage is applied to the full winding; the output is taken from a part of the winding.

(5)

a) Design an RL low pass filter that uses a 30-mH coil and has a cutoff frequency of 5 kHz.

(a)

To design an RL low-pass filter:

1. Formula for cutoff frequency ($f_c$):

   $$f_c = \frac{R}{2 \pi L}$$

2. Given values:

   • Inductance ($L$) = 30 mH = 0.03 H
   • Cutoff frequency ($f_c$) = 5 kHz = 5000 Hz

3. Find Resistance ($R$):
   Rearrange the formula:

   $$R = 2 \pi f_c L$$

   Substituting:

   $$R = 2 \pi (5000) (0.03) = 942 \, \Omega$$

Answer:

The RL low-pass filter requires:

• Inductor ($L$) = 30 mH
• Resistor ($R$) = 942 Ω

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