Explain the reason why direct current (DC) controller can be called as a counter emf controller.
Answer to Problem 1SQ
DC controller can be called as a counter emf controller because the counter emf of the armature is used for controlling the motor acceleration.
Explanation of Solution
In DC motor, the generated counter electromotive force is proportional to the speed of the motor for constant field. The counter emf can differ from the applied voltage only by armature drop.
DC controller can automatically adjust the starting time intervals, based upon the connected load. If heavy load is connected, it has more inertia to overcome; therefore, long time is required to generate the counter emf. When the DC motor starts, counter emf produced across the armature is very low. As the acceleration of motor increases, the counter emf increases. If the voltage value across the armature reaches a particular value, the starting resistance can be reduced at the correct time by actuating the relay. Therefore, the counter emf is mainly responsible to control the acceleration of motor.
Conclusion:
Thus, the DC controller can be called as a counter emf controller because the counter emf of the armature is used for controlling the motor acceleration.
Want to see more full solutions like this?
Chapter 46 Solutions
Electric Motor Control
- A conductor 300 mm long carries a current of 13A and is at right-angles to a magnetic fieldbetween two circular pole faces, each of diameter 80 mm. If the total flux between the polefaces is 0.75 mWb, calculate the force exerted on the conductor. [ANS = 0.582 N]arrow_forwarda) find Rthb) Find Vth in the circuit c)Draw the Thevenin Equivalent of the circuit to tge left of the a and b terminalsarrow_forwardAn electric car runs on batteries, but needs to make constant stops to re-charge. If a trailer is attached to the car that carries a generator, and the generator is turned by a belt attached to the wheels of the trailer, will the car be able to drive forever without stopping?arrow_forward
- A singl core cable of voltage 30 kv. The diameter of Conductor is 3 cm. The diameter of cable is 25 cm. This cable has Two layer of insulator having arelative permittivity 5-3 respectively of The ratio of maximum electric stress of maximum electric stress 8 First layer to the of second layer is 10 Find & 1- The thickness of each layers. 3- The voltage of each layers. §. Layers The saving in radius of cable if another ungrading cable has the Same maximum electric stress, Total village, Conductor diameter of grading cable.arrow_forward66 KV sing care Cable has a drameter of conductor of 3 cm. The radius of cable is 10 cm. This Cable house Two relative permmitivity of insulation 6 and 4 respectively. If The ratio of maximum electric stress of first layer to the maximum eledric streep & second layer is s 1- find the village & each layers. 2- Min- electric stress J Cable 3- Compare the voltage of ungrading Cable has the same distance and relectric stresses.arrow_forwardPrelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.arrow_forward
- Solve it in a different way than the previous solution that I searched forarrow_forwardA lossless uncharged transmission line of length L = 0.45 cm has a characteristic impedance of 60 ohms. It is driven by an ideal voltage generator producing a pulse of amplitude 10V and width 2 nS. If the transmission line is connected to a load of 200 ohms, sketch the voltage at the load as a function of time for the interval 0 < t < 20 nS. You may assume that the propagation velocity of the transmission is c/2. Answered now answer number 2. Repeat Q.1 but now assume the width of the pulse produced by the generator is 4 nS. Sketch the voltage at the load as a function of time for 0 < t < 20 nS.arrow_forwardSolve this experiment with an accurate solution, please. Thank you.arrow_forward
Electricity for Refrigeration, Heating, and Air C...Mechanical EngineeringISBN:9781337399128Author:Russell E. SmithPublisher:Cengage Learning