A DC cumulative compounded motor delivers rated load torque at rated speed. If the series field is short circuited, then the armature current and speed will be ?
Above is the question I'm trying to solve. Let me explain it first. Below is the figure of a cumulative compounded motor.
I'm using the following notations:
$\omega_m$ = mechanical speed of rotor in radian per sec(mechanical) = a variable
$k$ = a constant
$\phi$ = $f(N_fI_f+N_{se}I_a)$ = a variable
$V$ = terminal voltage = constant
$I_a$ = armature current = a variable
$R_a$ = armature resistance
$R_{se}$ = series field resistance
So the equation of consideration is $$ \omega_m = \frac{V}{k\phi}-\frac{I_a}{k\phi}(R_a+R_{se}) $$
So I've partially differentiated the above equation to get the following two equations.
$$
\frac{\partial \omega_m}{\partial R_{se}} = -\frac{I_a}{k\phi}R_a
$$
$$
\frac{\partial I_a}{\partial R_{se}} = -\frac{R_a(V-\omega_m k\phi)}{R_a+R_{se}}
$$
So, as mentioned in the question the $R_{se}$ has to made short circuited,i.e, $R_{se}=0$. So, what can I say about the nature of $\omega_m$ and $I_a$. Since the slope is coming negative I'm guessing that both the value will increase. Do I need to do partial derivative w.r.t. $\phi$ also because it's a variable, to come to the final conclusion? I'm getting confused in using partial derivative in such cases.