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+### Statement
+
+$11.1.29.$ [Insert the problem statement]
+
+### Solution
+
+
+For the first circuit we have:
+
+
+\begin{equation}
+\mathcal{E}+\mathcal{E}_{i1}=IR
+\end{equation}
+
+where $\mathcal{E}_{i1}$ is an induced emf of a motor in the first case.
+
+
+
+
+
+
+
+Similarly,for the second one we could write:
+
+
+\begin{equation}
+\mathcal{E}_0+\mathcal{E}_{i2}=0
+\end{equation}
+
+
+where $\mathcal{E}_{i2}$ is an induced emf of a motor in the second circuit.
+
+
+
+
+Using the fact the induced emf is proportional to its frequency we get:
+
+\begin{equation}
+\frac{\mathcal{E}_{i1}}{f}=\frac{\mathcal{E}_{i2}}{f_0}
+\end{equation}
+
+
+Or it is the same as:
+\begin{equation}
+\mathcal{E}_{i1}=-\mathcal{E}_0\frac{f}{f_0}
+\end{equation}
+
+
+In the steady state, we know that the net torque is zero,so:
+
+
+
+\begin{equation}
+M+M_m=0
+\end{equation}
+
+
+where $M_m$ is an electromagnetic torque of the motor which we could through it's angular frequency and power $P_m$:
+
+
+\begin{equation}
+\omega M_m=P_m
+\end{equation}
+
+
+
+
+But we could write $P_m$ as $\mathcal{E}_{i1}I$ and $\omega$ as $2\pi f$ for the first case.By plugging them into eq(6) and using eq(5) we get $I$:
+
+
+\begin{equation}
+I=\frac{2\pi M f_0}{\mathcal{E}_0}
+\end{equation}
+
+
+
+Finally,plugging last equation into the first one and using eq(4) we get our answer for frequency:
+
+
+\begin{equation}
+f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2})
+\end{equation}
+
+#### Answer
+
+[Insert a concise answer or boxed result]