Additionally loss due to brush contact resistance enclosed in armature copper loss
Loss within the armature due its rotation within the magnetic field of the poles.
3-15%
Friction: in bearings, brushes, etc.
Windage: air friction of rotating coil
Losses rely on the speed of the machine.
1% of FL Output Power
Miscellaneous losses occurs due to
Distortion of flux because of armature reaction
Short circuit currents in the coil, undergoing commutation
Very difficult to determine a reasonable value is assigned
Iron losses
Mechanical losses
Shunt field losses
Copper loss in armature
Copper loss in series field
\[\begin{aligned} \mbox{Generator output} & =VI\\ \mbox{Generator input} & =\mbox{output+losses} =VI+I_{a}^{2}R_{a}+W_{c}\\ & =VI+I_{a}^{2}R_{a}+W_{c}\\ & =VI+\left(I+I_{sh}\right)^{2}R_{a}+W_{c} \end{aligned}\] \(I_{sh}\) is negligible as compared to load current, then \(I_a=I\) \[\begin{aligned} \eta & =\dfrac{\mbox{output}}{\mbox{input}} = \dfrac{VI}{VI+I_{a}^{2}R_{a}+W_{c}}\\ & =\dfrac{VI}{VI+I^{2}R_{a}+W_{c}}\left(\because I_{a}=I\right)\\ & =\dfrac{1}{1+\left(\dfrac{IR_{a}}{V}+\dfrac{W_{c}}{VI}\right)} \end{aligned}\]
Now, efficiency is maximum when denominator is minimum i.e. when \[\begin{aligned} \dfrac{d}{dI}\left(\dfrac{IR_{a}}{V}+\dfrac{W_{c}}{VI}\right) & = 0\\ \dfrac{R_a}{V} - \dfrac{W_c}{VI^2} & = 0\\ I^2R_a & = W_c \end{aligned}\] Hence, generator efficiency is maximum when \[\boxed{\mbox{Variable loss = constant loss}}\] The load current corresponding to maximum efficiency is \[\begin{aligned} I^2R_a & = W_c\\ &\boxed{I = \sqrt{\dfrac{W_c}{R_a}}} \end{aligned}\]