For the system in the diagramD. The line current at startup. Use this per-phase equivalent circuit:Apply Ohm's law, note that slip is equal to 1 at startup and the resulting current is complex. Show the current in polar form.E. The starting torque. For each phase, Tstart = Pin/ധs , and the input power is the "line current squared times the equivalent rotor resistance given as 0.25/slip with slip = 1 at startup."F. Find the full-load current with a star starter and the power factor. Use Ifl = VL/Zeq, VL from the above circuit, and Zeq must include the slip calculated in problem 4. Show the complex current in polar formG. Determine the full-load torque. The suitable formula for the full-load torque is derived as:Tfl = (per phase x 3) Ifl2 x Req / (ധs x slip). Req is the sum of all the resistive impedances from the above circuit. Get the synchronous speed, and slip from problem 4. Note: From Problem 4:Synchronous Speed Ns = 1800 rpmSlip s = 2.5% The diagram presents an electrical circuit consisting of several components arranged in a combination of series and parallel connections. ### Components and Connections:1. **Voltage Source:** - The circuit is powered by a voltage source of \( \frac{480}{\sqrt{3}} \) V.2. **Resistors and Inductors:** - The top branch contains three impedances in series: - A resistor of 0.15 Ω. - An impedance of 0.9 Ω. - An impedance of 1.3 Ω.3. **Parallel Branch:** - There is a parallel connection between points A and B consisting of: - A 50 Ω resistor. - A 21 Ω resistor.4. **Right Branch:** - In series with the parallel combination, there is another impedance represented by: - An impedance of \( \frac{0.25}{s} \) Ω.### Explanation:- The top branch resistors and impedances are connected in series, meaning the same current flows through each component.- The resistors of 50 Ω and 21 Ω are connected in parallel between points A and B, so the voltage across each resistor is the same but the total current is the sum of currents through each resistor.- The total impedance of the parallel branch needs to be calculated considering the individual resistances and the subsequent resistors in series in the right branch.- The right branch containing the impedance of \( \frac{0.25}{s} \) Ω is typically a frequency-dependent component, possibly an inductor or capacitor in the s-domain (Laplace transform).This configuration is a typical example found in alternating current circuit analysis, where impedance and phasor notations are commonly used to simplify the calculations.

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Introductory Circuit Analysis (13th Edition)

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For the system in the diagram

D. The line current at startup. Use this per-phase equivalent circuit:
Apply Ohm's law, note that slip is equal to 1 at startup and the resulting current is complex. Show the current in polar form.

E. The starting torque. For each phase, T_start = P_in/ധ_s , and the input power is the "line current squared times the equivalent rotor resistance given as 0.25/slip with slip = 1 at startup."

F. Find the full-load current with a star starter and the power factor. Use I_fl = V_L/Z_eq, V_L from the above circuit, and Z_eq must include the slip calculated in problem 4. Show the complex current in polar form

G. Determine the full-load torque. The suitable formula for the full-load torque is derived as:

T_fl = (per phase x 3) I_fl² x R_eq / (ധ_s x slip). R_eq is the sum of all the resistive impedances from the above circuit. Get the synchronous speed, and slip from problem 4.

Note: From Problem 4:

Synchronous Speed N_s = 1800 rpm

Slip s = 2.5%

The diagram presents an electrical circuit consisting of several components arranged in a combination of series and parallel connections.

### Components and Connections:

1. **Voltage Source:**
- The circuit is powered by a voltage source of \( \frac{480}{\sqrt{3}} \) V.

2. **Resistors and Inductors:**
- The top branch contains three impedances in series:
- A resistor of 0.15 Ω.
- An impedance of 0.9 Ω.
- An impedance of 1.3 Ω.

3. **Parallel Branch:**
- There is a parallel connection between points A and B consisting of:
- A 50 Ω resistor.
- A 21 Ω resistor.

4. **Right Branch:**
- In series with the parallel combination, there is another impedance represented by:
- An impedance of \( \frac{0.25}{s} \) Ω.

### Explanation:

- The top branch resistors and impedances are connected in series, meaning the same current flows through each component.
- The resistors of 50 Ω and 21 Ω are connected in parallel between points A and B, so the voltage across each resistor is the same but the total current is the sum of currents through each resistor.
- The total impedance of the parallel branch needs to be calculated considering the individual resistances and the subsequent resistors in series in the right branch.
- The right branch containing the impedance of \( \frac{0.25}{s} \) Ω is typically a frequency-dependent component, possibly an inductor or capacitor in the s-domain (Laplace transform).

This configuration is a typical example found in alternating current circuit analysis, where impedance and phasor notations are commonly used to simplify the calculations.

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