Решение.
Запишем закон Ома для полной цепи:
\[ \begin{align}
& I=\frac{\xi }{R+r},\xi ={{I}_{1}}\cdot {{R}_{1}}+{{I}_{1}}\cdot r\ \ \ (1),\ \xi ={{I}_{2}}\cdot {{R}_{2}}+{{I}_{2}}\cdot r\ \ \ (2).\ P={{I}^{2}}\cdot R,\ {{R}_{1}}=\frac{{{P}_{1}}}{I_{1}^{2}}\ \ \ (3),{{R}_{2}}=\frac{{{P}_{2}}}{I_{2}^{2}}\ \ \ (4), \\
& \xi ={{I}_{1}}\cdot \frac{{{P}_{1}}}{I_{1}^{2}}+{{I}_{1}}\cdot r,\ \xi =\frac{{{P}_{1}}}{{{I}_{1}}}+{{I}_{1}}\cdot r\ \ \ (5),\ \xi ={{I}_{2}}\cdot \frac{{{P}_{2}}}{I_{2}^{2}}+{{I}_{2}}\cdot r,\ \xi =\frac{{{P}_{2}}}{I_{2}^{{}}}+{{I}_{2}}\cdot r\ \ \ (6). \\
& \frac{{{P}_{1}}}{{{I}_{1}}}+{{I}_{1}}\cdot r\ =\frac{{{P}_{2}}}{I_{2}^{{}}}+{{I}_{2}}\cdot r\ ,\ {{I}_{1}}\cdot r-{{I}_{2}}\cdot r=\frac{{{P}_{2}}}{{{I}_{2}}}-\frac{{{P}_{1}}}{{{I}_{1}}},\ r=\frac{\frac{{{P}_{2}}}{{{I}_{2}}}-\frac{{{P}_{1}}}{{{I}_{1}}}}{{{I}_{1}}-{{I}_{2}}}\ \ \ (7). \\
& r=\frac{\frac{200}{10}-\frac{180}{30}}{30-10}=0,7.\ \xi =\frac{180}{30}+30\cdot 0,7=27. \\
\end{align} \]
r = 0,7 Ом,
ЭДС = 27 В.
