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:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.


For the circuit shown above, the equivalent gate is:
