---
title: "Two semi converters (50 N-m at 1000 rpm)"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:semi-converters-field-armature-50nm-1000rpm"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/semi-converters-field-armature-50nm-1000rpm
site: BCT Engineering
---

# Two semi converters (50 N-m at 1000 rpm)

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2079 Bhadra, 8 marks; 2072 Kartik (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/semi-converters-field-armature-50nm-1000rpm
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A single phase, 220 V, 50 Hz supply feeds a separately excited dc motor through two single phase semi-converters, one for the field and another for armature. The firing angle for the field semi-converter is zero. The field resistance is 250 Ω and armature resistance is 0.2 Ω. The load torque is 50 N-m at 1000 rpm. The voltage constant is 0.8 V/A-rad/s and the torque constant is 0.8 N/A². Assuming the armature and field currents to be continuous and neglecting losses, Determine:
i) Field current.
ii) Firing angle of the converter in the armature circuit.
iii) Power factor of the converter of armature.

## Short answer

Field current is Vf/Rf = 198.07/250 = 0.792 A. Armature firing angle is found from Va = Eg + IaRa = 82.15 V, using Va = (Vm/pi)(1+cos alpha), giving alpha = 99.8 degrees. Armature converter power factor is 0.559 lagging.

## Answer

**Data:** $V_s = 220$ V, $V_m = \sqrt2\times220 = 311.13$ V; $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$; $T = 50$ N-m at $N = 1000$ rpm; $K_v = 0.8$ V/A-rad/s, $K_t = 0.8$ N-m/A². For a single-phase semiconverter with continuous current:

$$
V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha)
$$

#### i) Field current

Field firing angle $\alpha_f = 0$:

$$
\begin{aligned}
V_f &= \frac{311.13}{\pi}(1 + 1) = 198.07\ \text{V} \\
I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A}
\end{aligned}
$$

#### ii) Armature firing angle

Armature current from $T = K_tI_fI_a$:

$$
I_a = \frac{50}{0.8\times0.7923} = 78.89\ \text{A}
$$

Speed $\omega = \dfrac{2\pi\times1000}{60} = 104.72$ rad/s. Back emf:

$$
E_g = K_v\,\omega\,I_f = 0.8\times104.72\times0.7923 = 66.37\ \text{V}
$$

Armature voltage:

$$
V_a = E_g + I_aR_a = 66.37 + 78.89\times0.2 = 82.15\ \text{V}
$$

Firing angle:

$$
\begin{aligned}
82.15 &= \frac{311.13}{\pi}(1 + \cos\alpha_a) \\
\cos\alpha_a &= \frac{82.15\pi}{311.13} - 1 = -0.1705 \\
\alpha_a &= 99.82^\circ
\end{aligned}
$$

#### iii) Power factor of armature converter

With ripple-free armature current $I_a$, the supply current is a quasi-square wave of height $I_a$ conducting for $(\pi - \alpha)$ in each half cycle.

$$
\begin{aligned}
I_s &= I_a\sqrt{\frac{\pi-\alpha}{\pi}} = 78.89\sqrt{\frac{180-99.82}{180}} = 52.65\ \text{A} \\
I_{s1} &= \frac{2\sqrt2}{\pi}I_a\cos\frac{\alpha}{2} = 0.9003\times78.89\times0.6440 = 45.74\ \text{A}
\end{aligned}
$$

Displacement factor $= \cos(\alpha/2) = \cos 49.91^\circ = 0.644$.

$$
PF = \frac{I_{s1}}{I_s}\cos\frac{\alpha}{2} = \frac{45.74}{52.65}\times0.644 = 0.559\ \text{(lagging)}
$$

Check by power balance (losses neglected): $PF = \dfrac{V_aI_a}{V_sI_s} = \dfrac{82.15\times78.89}{220\times52.65} = 0.559$.

**Answer:** (i) $I_f = 0.792$ A; (ii) $\alpha_a = 99.8^\circ$; (iii) PF $= 0.559$ lagging.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With field firing angle 0, If is:
   A. 0.792 A
   B. 0.88 A
   C. 1.24 A
   D. 0.4 A
   Answer: A. Vf = (311.13/pi)(2) = 198.07 V; If = 198.07/250 = 0.792 A. (Key point: Field current)

2. Ia is found from:
   A. T = Kt If Ia, giving 78.89 A
   B. T = Kv If
   C. Ia = V/Ra
   D. T = Ia Ra
   Answer: A. Ia = 50/(0.8 x 0.7923) = 78.89 A. (Key point: Armature current)

3. Va equals:
   A. Eg only
   B. Eg + Ia Ra = 82.15 V
   C. Eg - Ia Ra
   D. Vs
   Answer: B. Eg = 66.37 V, Ia Ra = 15.78 V, so Va = 82.15 V. (Key point: Armature voltage)

4. Using Va = (Vm/pi)(1 + cos alpha), alpha is about:
   A. 45°
   B. 60°
   C. 99.8°
   D. 20°
   Answer: C. cos alpha = -0.1705 so alpha = 99.8°. (Key point: Firing angle)

## Close variants set in other exams

- [Two semi converters for field and armature](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Asemi-converters-field-armature-firing-angle) — Utilization of Electrical Energy, chapter 3, asked 3×

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