In the circuit shown, two batteries with emfs $e_1$ and $e_2$ are connected in series with resistors $R_1$ and $R_2$. If the current flowing through $e_2$ is $i_2$, what is its value?

• A. $i_2=\frac{e_2-e_1}{R_1+R_2}$
• B. $i_2=\frac{e_1+e_2}{R_1+R_2}$
• C. $i_2=\frac{e_1-e_2}{R_1+R_2}$
• D. $i_2=\frac{e_2}{R_2}$

Answer: (B)

Verified step by step guidance

1. Identify that the two batteries with emfs \(e_1\) and \(e_2\) are connected in series with resistors \(R_1\) and \(R_2\). Since they are in series, the same current \(i_2\) flows through both batteries and resistors.
2. Apply Kirchhoff's Voltage Law (KVL) around the loop, which states that the sum of the emfs is equal to the sum of the voltage drops across the resistors. Write the equation as: \(e_1 + e_2 = i_2 (R_1 + R_2)\).
3. Rearrange the equation to solve for the current \(i_2\): \(i_2 = \frac{e_1 + e_2}{R_1 + R_2}\).
4. Interpret the result: the total emf in the circuit is the algebraic sum of the two emfs, and the total resistance is the sum of the two resistors since they are in series.
5. Conclude that the current \(i_2\) flowing through the battery \(e_2\) is given by the total emf divided by the total resistance, as expressed in the formula derived.