
The armature of the dc motor has \[20\Omega \] resistance. It draws current of 1.5 ampere when run by 220 volts dc supply. The value of back e.m.f. induced in it will be
A) 190 V
B) 150 V
C) 170 V
D) 120 V
Answer
574.8k+ views
Hint:Back e.m.f. is the counter or opposite of e.m.f. therefore, it is also known as counter e.m.f. The Fleming right-hand rule can be used to determine the direction of the e.m.f. induced. According to Fleming Right Hand rule, if you hold your thumb, middle finger and index finger of the right hand by an angle of ${90^ \circ }$ , then the index finger will represent the direction of the magnetic field, the thumb in the direction of the motion and the middle finger, in the direction of e.m.f. induced.
Complete step by step:
What is back e.m.f. ?
When the armature of a direct current (d.c.) motor rotates under the influence of the torque, the armature conductors will move through the magnetic field and therefore, e.m.f. induced in them. Now, the em.f. induced acts in the counter or opposite direction to the applied voltage V (Lenz’s law) and it is known as counter or back e.m.f.
Given in the question,
Resistance in the d.c. motor $(R) = 20\Omega $
Current drawn $(i) = 1.5A$
Voltage supplied $(E) = 220V$
Formula used:
Current induced for d.c. motor,
$i = \left( {\dfrac{{\left( {E - e} \right)}}{R}} \right)$ ...(1)
Where, $e$ is the required back e.m.f. induced
Now from equation (1),
$E - e = iR$
Or, $e = E - iR$
Now, on putting the values in the equation,
$e = 220 - 1.5 \times 20$
Or, \[e = 190V\]
Hence, the correct answer is A. 190 V.
Note:Back e.m.f. is a type of system that is present in the coil of an electric motor which opposes the current flow through the coil, the armature starts rotating. If there is variation in the speed, there may be some change in the winding characteristics, which will result in variation in the back e.m.f. also. At high speeds, the back e.m.f. is high due to the rapid rotation of the armature.
Complete step by step:
What is back e.m.f. ?
When the armature of a direct current (d.c.) motor rotates under the influence of the torque, the armature conductors will move through the magnetic field and therefore, e.m.f. induced in them. Now, the em.f. induced acts in the counter or opposite direction to the applied voltage V (Lenz’s law) and it is known as counter or back e.m.f.
Given in the question,
Resistance in the d.c. motor $(R) = 20\Omega $
Current drawn $(i) = 1.5A$
Voltage supplied $(E) = 220V$
Formula used:
Current induced for d.c. motor,
$i = \left( {\dfrac{{\left( {E - e} \right)}}{R}} \right)$ ...(1)
Where, $e$ is the required back e.m.f. induced
Now from equation (1),
$E - e = iR$
Or, $e = E - iR$
Now, on putting the values in the equation,
$e = 220 - 1.5 \times 20$
Or, \[e = 190V\]
Hence, the correct answer is A. 190 V.
Note:Back e.m.f. is a type of system that is present in the coil of an electric motor which opposes the current flow through the coil, the armature starts rotating. If there is variation in the speed, there may be some change in the winding characteristics, which will result in variation in the back e.m.f. also. At high speeds, the back e.m.f. is high due to the rapid rotation of the armature.
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