Electric Motor Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
A DC motor has a back-EMF of 10 V10 \text{ V} when running at full speed. The supply voltage is 12 V12 \text{ V} and the coil resistance is 0.5 Ω0.5 \text{ } \Omega. What is the running current, the starting current (back-EMF = 0), and the mechanical power output?

Solution

  1. 1
    Running current: I=V−EbackR=12−100.5=20.5=4 AI = \frac{V - \mathcal{E}_{\text{back}}}{R} = \frac{12 - 10}{0.5} = \frac{2}{0.5} = 4 \text{ A}.
  2. 2
    Starting current (no back-EMF): Istart=VR=120.5=24 AI_{\text{start}} = \frac{V}{R} = \frac{12}{0.5} = 24 \text{ A} — much higher!
  3. 3
    Mechanical power: Pmech=Eback×I=10×4=40 WP_{\text{mech}} = \mathcal{E}_{\text{back}} \times I = 10 \times 4 = 40 \text{ W}. Power wasted as heat: Pheat=I2R=16×0.5=8 WP_{\text{heat}} = I^2R = 16 \times 0.5 = 8 \text{ W}.

Answer

Irun=4 A,  Istart=24 A,  Pmech=40 WI_{\text{run}} = 4 \text{ A}, \; I_{\text{start}} = 24 \text{ A}, \; P_{\text{mech}} = 40 \text{ W}
A spinning motor generates a back-EMF that opposes the supply voltage, reducing the current. At startup there is no back-EMF, so the current is very high — this is why motors need starting resistors or soft-start circuits.

About Electric Motor

A device that converts electrical energy into mechanical energy (rotational motion) by exploiting the force exerted on a current-carrying conductor placed in a magnetic field.

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Example 2 medium

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Example 3 medium

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