DC shunt motor is coupled to the identical DC shunt generator. The fi...
Motor is connected across rated voltage V = 1 pu
Rated armature current flaws in bath the motor and generator Im=Ig=1pu
Back emf in motor
Eb = V − ImRa = 1 − 1 × 0.02 = 0.98pu
Mechanical output power of motor
= EbIm -mechanicallasses
= 0.98 × 1 − 0.05 = 0.93 pu
This power is given to the generator.
Output power of generator = Output power of motor - mechanical Iasses
EgIg = 0.93 − 0.05 = 0.88pu ⇒ Eg = 0.88puTerminal voltage of generator
= Eg − lgRa = 0.88 − 0.02 = 0.86pu
Laad resistance
RL= V9/I9 = 0.86/1 = 0.86pu
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DC shunt motor is coupled to the identical DC shunt generator. The fi...
To determine the load resistance across the generator in order to obtain the rated armature current in both the motor and generator, we need to analyze the circuit and calculate the equivalent resistance.
1. Equivalent circuit:
The equivalent circuit of the coupled motor-generator system can be represented as follows:
- The motor armature resistance is given as 0.02 pu (per unit).
- The mechanical losses are given as 0.05 pu.
- The armature resistance of the generator is RL (unknown).
- The field of the generator is connected to the same supply source as the motor.
2. Rated voltage and current:
Since the rated voltage is applied across the motor, we can assume that the voltage across the generator is also the rated voltage. This implies that the rated armature current in both the motor and generator is the same.
3. Calculation:
Using the equivalent circuit, we can calculate the total voltage drop across the armature resistance and mechanical losses in the motor. This can be expressed as:
Voltage drop in motor = Rated voltage - Voltage across generator = Rated voltage - Rated voltage = 0
Since the voltage drop across the motor is zero, the total resistance in the motor circuit (armature resistance + mechanical losses) must also be zero. Therefore, we can write the following equation:
0.02 pu (motor armature resistance) + 0.05 pu (mechanical losses) = 0
Simplifying the equation, we get:
0.02 + 0.05 = 0
0.07 pu = 0
This equation is not possible, as it leads to an inconsistency. Therefore, we can conclude that the motor and generator cannot have the same rated armature current.
4. Answer:
Since it is not possible to obtain the same rated armature current in both the motor and generator, the given question is not solvable. Therefore, none of the provided options (a, b, c, d) is correct.