From the given current equations, we can cross-multiply:
\[ I_1 (R_1 + r) = E \quad \text{and} \quad I_2 (R_2 + r) = E \]
Equating the two equations for \( E \), we get:
\[ I_1 (R_1 + r) = I_2 (R_2 + r) \]
Expanding both sides:
\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]
Rearranging the terms:
\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \]
Factor out \( r \):
\[ r (I_1 - I_2) = I_2 R_2 - I_1 R_1 \]
Solving for \( r \), we get:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
The internal resistance \( r \) is given by:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: