Question

Calculate speed of motor when delivery maximum voltage. Also calculate the resistance to be added to achieve 3/4th of maximum torque at time of starting.

Answer 1

To solve this, we need more information about the motor, such as its type (DC or AC), ratings, and circuit parameters. Let's assume we're dealing with a **DC motor**, as they are commonly used in such calculations.

### 1. **Speed of Motor at Maximum Voltage**
For a DC motor, the speed \( N \) is proportional to the applied voltage \( V \) and inversely proportional to the back EMF and armature resistance. The general equation for a DC motor's speed is:

\[
N = \frac{V - I_a R_a}{k}
\]

Where:
- \( V \) is the applied voltage.
- \( I_a \) is the armature current.
- \( R_a \) is the armature resistance.
- \( k \) is a constant based on motor characteristics.

At maximum voltage, the speed will be the highest, assuming no load conditions, and the current drawn will be minimal. In such a case, the back EMF is almost equal to the applied voltage, and the speed can be approximated as:

\[
N_{max} = \frac{V_{max}}{k}
\]

But to provide an accurate number, you would need to know the values of \( V_{max} \) and \( k \).

---

### 2. **Resistance to Achieve 3/4th of Maximum Torque at Starting**

The starting torque of a DC motor is proportional to the current through the armature:

\[
T \propto I_a
\]

For the torque to be 3/4th of the maximum torque, we need to adjust the armature current. The current at startup (when back EMF is zero) is given by:

\[
I_a = \frac{V}{R_a + R_{ext}}
\]

Where \( R_{ext} \) is the external resistance to be added. To achieve 3/4th of the maximum torque, the current should be reduced to 3/4th of the original starting current:

\[
\frac{I_{new}}{I_{start}} = \frac{3}{4}
\]

Thus:

\[
\frac{V}{R_a + R_{ext}} = \frac{3}{4} \cdot \frac{V}{R_a}
\]

Simplifying:

\[
R_a + R_{ext} = \frac{4}{3} R_a
\]

\[
R_{ext} = \frac{4}{3} R_a - R_a = \frac{R_a}{3}
\]

So, the external resistance \( R_{ext} \) to be added should be one-third of the armature resistance \( R_a \).

---

For exact values, you would need:
- Maximum voltage \( V_{max} \)
- Armature resistance \( R_a \)
- Motor constant \( k \)

Answer 2

To solve these problems, you’ll need to work with some electrical and mechanical principles related to electric motors, specifically DC motors or induction motors. Here’s a step-by-step guide to addressing both parts of your query:

### 1. Calculate Speed of Motor When Delivering Maximum Voltage

For a DC motor, the speed \( N \) of the motor is directly proportional to the applied voltage \( V \) and inversely proportional to the motor's field flux \( \Phi \). The relationship is given by:

\[ N = \frac{V - I_a R_a}{k \Phi} \]

Where:
- \( V \) = applied voltage
- \( I_a \) = armature current
- \( R_a \) = armature resistance
- \( k \) = a constant depending on motor construction
- \( \Phi \) = magnetic flux per pole (related to the field current)

When delivering maximum voltage, the speed will be at its maximum if the armature current is minimized. Assuming that the field flux \( \Phi \) is constant, the speed will be:

\[ N_{max} = \frac{V_{max}}{k \Phi} \]

**In practice:** The exact speed depends on the specifics of the motor design and the values of the parameters. If you have the maximum voltage and the motor's constants, you can calculate it directly using the formula above.

### 2. Calculate the Resistance to Achieve 3/4th of Maximum Torque at Startup

For DC motors, the torque \( T \) is proportional to the armature current \( I_a \) and the magnetic flux \( \Phi \):

\[ T = k \Phi I_a \]

At startup, the torque is maximum, and as the motor speeds up, the torque decreases. To achieve 3/4th of the maximum torque at startup, you need to adjust the resistance in the armature circuit.

The torque at startup \( T_{start} \) is given by:

\[ T_{start} = k \Phi I_{a,start} \]

The desired torque \( T_{desired} \) at 3/4th of maximum torque is:

\[ T_{desired} = \frac{3}{4} T_{max} \]

At startup, the armature current is maximum. To achieve the desired torque, you need to add a series resistance \( R_s \) such that the armature current is reduced. The new armature current \( I_{a,new} \) can be calculated as follows:

\[ I_{a,new} = \frac{I_{a,max}}{R_a + R_s} \]

Where \( R_a \) is the armature resistance, and \( R_s \) is the additional resistance.

To calculate \( R_s \), follow these steps:

1. Determine the maximum torque condition. This typically means the initial armature current \( I_{a,max} \) at startup is used.
2. Calculate the desired armature current \( I_{a,desired} \) to produce 3/4th of the maximum torque. The current will be:

\[ I_{a,desired} = \frac{3}{4} I_{a,max} \]

3. Use Ohm’s law and the formula for the armature current to find the required series resistance \( R_s \):

\[ V_{armature} = I_{a,new} (R_a + R_s) \]

\[ R_s = \frac{V_{armature}}{I_{a,desired}} - R_a \]

Where \( V_{armature} \) is the applied voltage to the armature.

**In summary:**

1. To find the speed at maximum voltage, use the speed formula considering the maximum voltage and the constants of the motor.
2. To find the resistance needed to achieve 3/4th of the maximum torque at startup, calculate the new armature current needed and then determine the additional resistance required to reduce the current to that level.

If you have specific values for voltage, armature resistance, and other parameters, you can plug them into these formulas to get the precise results.