1. **Stating the problem:** We have a closed electrical circuit loop with voltage sources $v_a$, $v_b$ and resistors $R_1$, $R_2$, $R_3$ carrying current $i$. The equations given are: $$-v_a + v_1 + v_b + v_2 + v_3 = 0$$ $$-v_a + iR_1 + v_b + iR_2 + iR_3 = 0$$ $$v_b - v_a = i(R_1 + R_2 + R_3)$$ 2. **Understanding the problem:** This is an application of Kirchhoff's Voltage Law (KVL), which states that the sum of voltage changes around any closed loop is zero. 3. **Formula used:** The voltage across each resistor is given by Ohm's Law: $$v = iR$$ where $v$ is voltage, $i$ is current, and $R$ is resistance. 4. **Step-by-step explanation:** - The first equation sums all voltages around the loop and sets them to zero. - The second equation substitutes $v_1 = iR_1$, $v_2 = iR_2$, and $v_3 = iR_3$ into the first equation. - The third equation rearranges terms to isolate the voltage difference between $v_b$ and $v_a$. 5. **Simplifying the third equation:** $$v_b - v_a = i(R_1 + R_2 + R_3)$$ This means the voltage difference between the two sources equals the total voltage drop across all resistors. 6. **Solving for current $i$:** $$i = \frac{v_b - v_a}{R_1 + R_2 + R_3}$$ 7. **Summary:** - Use KVL to write the voltage sum equation. - Use Ohm's Law to express resistor voltages. - Rearrange to find current $i$. This approach helps analyze the circuit easily by breaking it into voltage sources and resistor drops.