---
title: "Starting current and starting resistance of shunt motor"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-220v-starting-current-80pc"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-starting-current-80pc
site: BCT Engineering
---

# Starting current and starting resistance of shunt motor

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2077 Chaitra, 8 marks
- Video: [DC Motor Starters](https://www.youtube.com/watch?v=TYUYIBG6YTA) — Electrical Educator
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-starting-current-80pc
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 220 V dc shunt motor has armature winding resistance of 0.08 ohm and field winding resistance of 100 ohms. Calculate the starting current drawn by the motor with normal input voltage switched on. If the starting current is to be reduced by 80%, calculate the value of starting resistance to be connected in series with armature circuit with input voltage reduced by 10%.

## Short answer

Starting armature current = 220/0.08 = 2750 A (line 2752.2 A). Desired current = 0.2×2750 = 550 A at 198 V. Required total resistance = 198/550 = 0.36 Ω, starter resistance = 0.36 – 0.08 = 0.28 Ω.

## Answer

At starting, speed is zero so back emf $E_b = 0$; only $R_a$ limits the armature current.

**Given:** $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 100\ \Omega$.

#### Part 1: Starting current at normal voltage
$$
\begin{aligned}
I_{a,st} &= \frac{V}{R_a} = \frac{220}{0.08} = 2750\ \text{A} \\
I_{sh} &= \frac{220}{100} = 2.2\ \text{A} \\
I_{L,st} &= 2750 + 2.2 = 2752.2\ \text{A}
\end{aligned}
$$

#### Part 2: Starting resistance
"Reduced by 80%" is read as: new starting armature current = 20% of 2750 A. Supply voltage is 10% lower, $V' = 0.9 \times 220 = 198$ V.

$$
I'_{a,st} = 0.2 \times 2750 = 550\ \text{A}
$$

$$
\begin{aligned}
I'_{a,st} &= \frac{V'}{R_a + R_{st}} \\
R_a + R_{st} &= \frac{198}{550} = 0.36\ \Omega \\
R_{st} &= 0.36 - 0.08 = 0.28\ \Omega
\end{aligned}
$$

(If line current is used instead: $I'_L = 0.2 \times 2752.2 = 550.44$ A, $I_{sh} = 1.98$ A, $I'_a = 548.46$ A, giving $R_{st} = 0.281\ \Omega$, practically the same.)

**Answer:** Starting current $\approx 2750$ A in the armature (2752.2 A from the line); starting resistance required $R_{st} \approx 0.28\ \Omega$.

## Check yourself

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

1. What limits starting armature current at normal voltage?
   A. Only Ra, since Eb = 0: Ia = V/Ra = 2750 A
   B. Ra + Rsh
   C. Rsh only
   D. The load torque
   Answer: A. 220/0.08. (Key point: Start current)

2. How is 'reduced by 80%' applied?
   A. 80% of 2750 A = 2200 A
   B. New starting armature current = 20% of 2750 A = 550 A
   C. 2750 - 80 A
   D. 550 V
   Answer: B. Reduction by 80% leaves 20%. (Key point: Interpretation)

3. What supply voltage is used in part 2?
   A. 242 V
   B. 220 V
   C. 198 V
   D. 176 V
   Answer: C. 10% lower than 220 V. (Key point: Voltage)

4. What starting resistance results?
   A. 0.36 Ω
   B. 0.08 Ω
   C. 0.44 Ω
   D. About 0.28 Ω
   Answer: D. 198/550 = 0.36 total, minus 0.08. (Key point: Resistance)

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