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.
1 Answer
To calculate the speed of the motor when delivering the maximum voltage and the resistance to be added to achieve 3/4th of maximum torque at the time of starting, we need to break down the problem into two parts.
Part 1: Speed of the motor when delivering maximum voltage
The speed of a motor depends on the applied voltage, the back EMF (electromotive force), and the motor's construction. For a DC motor, the speed \( N \) is proportional to the applied voltage \( V \) and inversely proportional to the field current. When the motor delivers maximum voltage, the back EMF \( E_b \) is at its maximum.
Let's use the following relationships:
At maximum voltage delivery, we assume the motor is running at its maximum speed and delivering the maximum voltage. The speed can be calculated based on the maximum voltage \( V_{max} \), and by knowing the motor's characteristics (such as the constant \( k \) and armature resistance \( R \)), you can estimate the speed. However, more specific information about the motor is needed to provide an exact formula or result.
Part 2: Resistance to be added to achieve 3/4th of maximum torque at starting
The torque \( T \) in a DC motor is given by the equation:
\[ T \propto I \cdot \Phi \]
Where \( I \) is the armature current, and \( \Phi \) is the magnetic flux (which depends on the field current). When the motor starts, the current \( I_{\text{start}} \) is at its maximum, and it decreases as the motor speeds up.
To achieve 3/4th of the maximum torque at starting, we need to limit the starting current, as the starting torque is proportional to the current. We can achieve this by adding resistance in series with the motor.
The torque is proportional to the square of the armature current, so if you want to achieve 3/4th of the maximum torque, you need to reduce the starting current by a factor of \( \sqrt{\frac{3}{4}} \).
The current reduction factor is:
\[
\frac{I_{\text{new}}}{I_{\text{start}}} = \sqrt{\frac{3}{4}} = 0.866
\]
So, the added resistance \( R_{\text{add}} \) should be chosen such that the current decreases by this factor.
Using Ohm’s Law:
\[
I_{\text{new}} = \frac{V}{R_{\text{total}}}
\]
Where \( R_{\text{total}} \) is the sum of the motor’s internal resistance \( R \) and the added resistance \( R_{\text{add}} \). By adjusting \( R_{\text{add}} \), we can achieve the desired starting current that will give 3/4th of the maximum torque.
To calculate the exact value of \( R_{\text{add}} \), you would need to know the motor’s rated voltage \( V \), armature resistance \( R \), and the starting current.
Summary:
If you have specific values for these parameters, I can help you calculate the exact numbers.
Part 1: Speed of the motor when delivering maximum voltage
The speed of a motor depends on the applied voltage, the back EMF (electromotive force), and the motor's construction. For a DC motor, the speed \( N \) is proportional to the applied voltage \( V \) and inversely proportional to the field current. When the motor delivers maximum voltage, the back EMF \( E_b \) is at its maximum.
Let's use the following relationships:
- \( E_b = k \cdot \Phi \cdot N \) (where \( k \) is a constant, \( \Phi \) is the flux, and \( N \) is the speed of the motor)
- \( E_b = V - I \cdot R \) (where \( V \) is the supply voltage, \( I \) is the armature current, and \( R \) is the armature resistance)
At maximum voltage delivery, we assume the motor is running at its maximum speed and delivering the maximum voltage. The speed can be calculated based on the maximum voltage \( V_{max} \), and by knowing the motor's characteristics (such as the constant \( k \) and armature resistance \( R \)), you can estimate the speed. However, more specific information about the motor is needed to provide an exact formula or result.
Part 2: Resistance to be added to achieve 3/4th of maximum torque at starting
The torque \( T \) in a DC motor is given by the equation:
\[ T \propto I \cdot \Phi \]
Where \( I \) is the armature current, and \( \Phi \) is the magnetic flux (which depends on the field current). When the motor starts, the current \( I_{\text{start}} \) is at its maximum, and it decreases as the motor speeds up.
To achieve 3/4th of the maximum torque at starting, we need to limit the starting current, as the starting torque is proportional to the current. We can achieve this by adding resistance in series with the motor.
The torque is proportional to the square of the armature current, so if you want to achieve 3/4th of the maximum torque, you need to reduce the starting current by a factor of \( \sqrt{\frac{3}{4}} \).
The current reduction factor is:
\[
\frac{I_{\text{new}}}{I_{\text{start}}} = \sqrt{\frac{3}{4}} = 0.866
\]
So, the added resistance \( R_{\text{add}} \) should be chosen such that the current decreases by this factor.
Using Ohm’s Law:
\[
I_{\text{new}} = \frac{V}{R_{\text{total}}}
\]
Where \( R_{\text{total}} \) is the sum of the motor’s internal resistance \( R \) and the added resistance \( R_{\text{add}} \). By adjusting \( R_{\text{add}} \), we can achieve the desired starting current that will give 3/4th of the maximum torque.
To calculate the exact value of \( R_{\text{add}} \), you would need to know the motor’s rated voltage \( V \), armature resistance \( R \), and the starting current.
Summary:
- Speed of the motor at maximum voltage: You need the motor's parameters (such as \( k \), \( R \), etc.) to calculate the exact speed.
- Resistance to add for 3/4th of maximum torque: Add resistance such that the current reduces by a factor of 0.866 to achieve 3/4th of maximum torque. The added resistance can be calculated using the voltage and the motor’s armature resistance.
If you have specific values for these parameters, I can help you calculate the exact numbers.