Question

Initially a d.c shunt motor having ra =0.5Ω and Rf=220Ω is running at 1000 rpm drawing 20 A from 220 V supply. If the field resistance is increased by 10%, calculate the new steady state armature current and speed of the motor. Assume the load torque to be constant.

Answer 1

The question requires an understanding of a DC shunt motor's operation, specifically the effects of increasing field resistance on motor speed and armature current. The increase in field resistance decreases field current and flux, leading to an increase in motor speed. The new armature current and speed depend on Ohm's Law and the motor's characteristics.

Step-by-step explanation:

The student's question involves a change in a direct current (DC) shunt motor's field resistance and the subsequent effects on the motor's speed and armature current, assuming constant load torque. The field resistance (Rf) is increased by 10%, changing from 220Ω to 242Ω (220Ω + 10% of 220Ω). As the field resistance increases, the field current decreases, resulting in a decrease in the field flux (Φ). Since the speed (N) of a DC shunt motor is inversely proportional to the field flux (N ∝ 1/Φ), the motor's speed will increase. The armature current (Ia) can be calculated using the supply voltage (V), armature resistance (Ra), and the back emf (Eback) which changes with speed.

Since the back emf is proportional to both the field flux and the speed (Eback ∝ ΦN), and the field flux decreases due to the increased resistance, the increased speed will result in a less pronounced increase in the back emf. The new steady state armature current can be found by applying Ohm's Law (I = V/R) and accounting for the new field resistance and the changes in the back emf. The exact calculation would require additional information on the motor's characteristics that is not provided in the question, such as the armature constant and the relationship between flux and armature current.

Answer 2

After increasing the field resistance by 10%, the new field resistance is 242Ω. Assuming constant load torque, the back emf and hence the armature current remain the same, and consequently, the steady-state armature current also remains at 19 A (20 A total minus 1 A field current). The motor speed will increase due to the reduced flux caused by the higher field resistance.

Step-by-step explanation:

When the field resistance Rf of a DC shunt motor is increased by 10%, the new field resistance becomes 1.1 × 220Ω = 242Ω. The initial armature current Ia was 20 A and the armature resistance Ra is given as 0.5Ω. The initial field current If is the current through Rf when the motor is connected to a 220 V supply: If = V / Rf = 220V / 220Ω = 1 A. The initial armature current is the total current minus field current: Ia = Itotal - If = 20 A - 1 A = 19 A The back emf Eb for the initial conditions can be found using: Eb = V - Ia * Ra = 220V - 19 A * 0.5Ω = 220V - 9.5V = 210.5V Since the motor speed is directly proportional to the back emf and inversely proportional to the flux, and the flux is inversely proportional to Rf, if Rf increases, the flux decreases, leading to an increase in speed. However, the question assumes constant load torque which implies that the back emf remains constant. Therefore, new armature current is calculated by assuming the back emf stays the same. With the new field resistance, the field current If' becomes: If' = V / Rf' = 220V / 242Ω = 0.909 A .The total armature current will now be: Ia' = Itotal' - If' = Itotal - If' .Itotal' can be found by using the back emf, which remains constant due to constant load torque: Eb = V - Ia' * Ra , 210.5V = 220V - Ia' * 0.5Ω , Ia' = (220V - 210.5V) / 0.5Ω = 19 A. The armature current remains the same as before the change in field resistance due to the assumptions provided. The motor speed will be higher than the initial 1000 rpm.