Irodov 3.193: Half the Input Turns the Shaft – Amit & Deepali Sinha Foundation

Problem Irodov 3.193

A direct-current electric motor runs from a voltage $V$ , and the resistance of its armature winding is $R$ . For what current through the winding is the useful (mechanical) power of the motor greatest? What is that greatest power, and what is the efficiency of the motor when it delivers it?

The motor M, whose armature winding has resistance R, running from the voltage V. — The motor $M$ , whose armature winding has resistance $R$ , running from the voltage $V$ .

Given

voltage $V$
winding resistance $R$

Find

the current

I

for the largest useful power · that power

P_{\mathrm{max}}

· the efficiency

\eta

then

Solution

Step 1Where the input power goes

The motor as a circuit: the source V, the winding resistance R, and the back emf E_i of the turning armature, which opposes the current I. — The motor as a circuit: the source $V$ , the winding resistance $R$ , and the back emf $\mathcal{E}_{\mathrm{i}}$ of the turning armature, which opposes the current $I$ .

The source delivers the power $VI$ to the motor. Part of it is Joule heat in the winding; the rest is the work done by the turning armature against its load, the useful power $P$ :

P = VI - I^2R \tag{1}

Step 2The current that maximises the useful power

The useful power (1) is a downward parabola in $I$ . Setting its derivative to zero:

\frac{dP}{dI} = V - 2IR = 0

I = \frac{V}{2R}

(2)

The second derivative, $-2R$ , is negative, so this is the maximum.

Step 3The greatest useful power

Substituting (2) into (1):

P_{\mathrm{max}} = V\cdot\frac{V}{2R} - \frac{V^2}{4R^2}\,R = \frac{V^2}{2R} - \frac{V^2}{4R}

P_{\mathrm{max}} = \frac{V^2}{4R}

(3)

The useful power against the current: zero with no current and zero again at I = V/R, where the armature stands still; the maximum lies halfway between. — The useful power against the current: zero with no current and zero again at $I = V/R$ , where the armature stands still; the maximum lies halfway between.

Step 4The efficiency at that current

The efficiency is the useful power over the input power $VI$ . At the current (2) the input is $V^2/2R$ , so

\eta = \frac{P_{\mathrm{max}}}{VI} = \frac{V^2/4R}{V^2/2R}

\eta = \frac{1}{2} = 50\,\%

(4)

The other half of the input is heat in the winding: $I^2R = V^2/4R$ , equal to the useful power.

Check

Dimensions: $V/R$ is a current and $V^2/R$ a power. Limits: with no current $P = 0$ , and at $I = V/R$ the whole voltage drops across the winding, the back emf is zero, the armature stands still and again $P = 0$ ; a parabola vanishing at $0$ and $V/R$ peaks at the midpoint $V/2R$ , as found.

Remark

This is the matching rule of a source and its load in disguise: the back emf $\mathcal{E}_{\mathrm{i}} = V/2$ at the optimum, so the armature behaves like a load resistance $\mathcal{E}_{\mathrm{i}}/I = R$ , equal to the winding’s own. A practical motor is run far from this point, at a current well below $V/2R$ , where the efficiency, $\mathcal{E}_{\mathrm{i}}/V$ in general, is close to one.