Problem 74 - Physics for Scientists and Engineers with Modern Physics

Question

The magnetic field of a "shunt-wound" dc motor is produced by field coils placed in parallel with the armature coils. Suppose that the field coils have a resistance of \(36.0 \Omega\) and the armature coils \(3.00 \Omega\). The back emf at full speed is \(105 \mathrm{~V}\) when the motor is connected to \(115 \mathrm{~V}\) dc. \((a)\) Draw the equivalent circuit for the situations when the motor is just starting and when it is running full speed. (b) What is the total current drawn by the motor at start up? \((c)\) What is the total current drawn when the motor runs at full speed?

Step-by-Step Solution

Answer

(b) Approximately 41.52 A at startup. (c) Approximately 6.52 A at full speed.

Step 1: Understand the Equivalent Circuit

For a shunt-wound DC motor, the field coils and armature coil are placed in parallel. The equivalent circuit can be depicted with a voltage source connected to two parallel branches: one containing the field coil with resistance \( R_f = 36.0 \Omega \) and the other containing the armature coil with resistance \( R_a = 3.00 \Omega \). Additionally, the back emf (\( E_b = 105 \mathrm{~V} \)) is considered across the armature coil when the motor is running at full speed.

Step 2: Analyze the Circuit at Startup

At startup, the motor has yet to generate any back emf (\( E_b = 0 \)). The total voltage \( V = 115 \mathrm{~V} \) is applied across both field and armature coils. Thus, the current in the armature (\( I_a \)) can be found using Ohm's law: \( I_a = \frac{V}{R_a} = \frac{115 \mathrm{~V}}{3.00 \Omega} \). The field coil current (\( I_f \)) is also calculated using Ohm's law: \( I_f = \frac{V}{R_f} = \frac{115 \mathrm{~V}}{36.0 \Omega} \).

Step 3: Calculate Total Current at Startup

The total current drawn from the source at startup \( I_{total,start} \) is the sum of currents through the armature and field coils: \( I_{total,start} = I_a + I_f \).

Step 4: Analyze the Circuit at Full Speed

When the motor is running at full speed, a back emf of \( E_b = 105 \mathrm{~V} \) is developed across the armature. The effective voltage across the armature coils is \( V_{armature} = V - E_b = 115 \mathrm{~V} - 105 \mathrm{~V} \). The armature current at full speed (\( I_{a, ext{full}} \)) is then calculated by \( I_{a, ext{full}} = \frac{V_{armature}}{R_a} \). The field coil current remains unchanged from the startup condition since the full voltage is still applied across it.

Step 5: Calculate Total Current at Full Speed

The total current drawn from the source at full speed \( I_{total,full} \) is the sum of the field coil current and the new armature current: \( I_{total,full} = I_{a, ext{full}} + I_f \).

Step 6: Conclude the Analysis

Perform the calculations to get \( I_{a} = \frac{115}{3} \approx 38.33 \mathrm{~A} \) and \( I_f = \frac{115}{36} \approx 3.19 \mathrm{~A} \). Hence, \( I_{total,start} \approx 41.52 \mathrm{~A} \). For full speed, \( I_{a, ext{full}} = \frac{10}{3} \approx 3.33 \mathrm{~A} \), so \( I_{total,full} \approx 6.52 \mathrm{~A} \).

Key Concepts

Equivalent circuit

In the discussion of shunt-wound DC motors, it's crucial to visualize the equivalent circuit. Picture this: a voltage supply provides power to two parallel branches. One branch consists of the field coils with a resistance of 36.0 Ohms, while the other branch houses the armature coils, which have a resistance of 3.00 Ohms.
Here’s a mental shortcut: the field coils and armature are like parallel lanes on a highway, with the voltage source as the starting point. When the motor is in operation, a back emf (electromotive force) also appears in the armature. This back emf is akin to an opposing force that must be reckoned with at full speed operation.
Visualizing the equivalent circuit is the first step to understanding how current flows and how different resistive forces come together to form an operational motor setup.

Ohm's law

Ohm's Law is a fundamental principle used throughout electronics to relate voltage, current, and resistance in a circuit. The formula is simple but powerful: \[ V = I \times R \] where \( V \) is the voltage across a component, \( I \) is the current through it, and \( R \) is its resistance.
For a shunt-wound DC motor at startup, Ohm's Law allows us to determine the current through both the armature and field coils. The total source voltage is 115 V. For the armature, the startup current \( I_a \) is \[ I_a = \frac{115 \text{ V}}{3.00 \Omega} \approx 38.33 \text{ A} \]
For the field coils, the calculation is \[ I_f = \frac{115 \text{ V}}{36.0 \Omega} \approx 3.19 \text{ A} \]
Using Ohm's Law enables us to quantify the flow of current and predict how much current each component must handle.

Back emf

A fascinating aspect of electric motors is the concept of back emf. As a motor reaches full speed, it generates its own electromotive force in the opposite direction of the applied voltage. This is known as the back emf.
In our shunt-wound DC motor, the back emf is 105 V when running at full speed. This means the effective voltage across the armature changes: \[ V_{armature} = 115 \text{ V} - 105 \text{ V} = 10 \text{ V} \]
This reduced effective voltage results in a different current flow through the armature.The back emf works as a natural regulator, preventing excessive current at high speeds, which can protect the motor from potential damage due to overheating.

Armature resistance

The armature resistance is a pivotal factor in determining how much current flows through the motor. It is a measure of how much the armature resists the flow of electric current.
For the shunt-wound DC motor, the armature resistance is 3.00 Ohms. At startup, this resistance, devoid of any back emf, allows a large current to flow: roughly 38.33 A.
When the motor is at full speed, the armature resistance helps to calculate the new current in conjunction with the back emf. With an effective voltage of only 10 V due to back emf, the armature current decreases significantly to \[ I_{a, \text{full}} = \frac{10 \text{ V}}{3.00 \Omega} \approx 3.33 \text{ A} \]
Understanding the role of armature resistance is key to predicting how the motor will behave under different operating conditions.