1. **Problem statement:** We have a series circuit with total resistance $R_{total} = 25\ \Omega$ and voltage drop across the battery $V = 6.0\ \text{V}$. We need to find: (a) The current in the circuit. (b) The voltage drop across each lamp. 2. **Formula used:** Ohm's Law: $$V = IR$$ where $V$ is voltage, $I$ is current, and $R$ is resistance. In a series circuit, the current $I$ is the same through all components. 3. **Step (a) Calculate the current:** Using Ohm's Law: $$I = \frac{V}{R_{total}} = \frac{6.0}{25}$$ $$I = 0.24\ \text{A}$$ 4. **Step (b) Calculate voltage drop across each lamp:** Since the lamps are in series, the total resistance is the sum of the resistances of the two lamps: $$R_{total} = R_1 + R_2$$ The problem does not specify individual resistances, so we assume both lamps have equal resistance: $$R_1 = R_2 = \frac{R_{total}}{2} = \frac{25}{2} = 12.5\ \Omega$$ Voltage drop across each lamp using Ohm's Law: $$V_1 = IR_1 = 0.24 \times 12.5 = 3.0\ \text{V}$$ $$V_2 = IR_2 = 0.24 \times 12.5 = 3.0\ \text{V}$$ **Final answers:** (a) Current in the circuit: $0.24\ \text{A}$ (b) Voltage drop across each lamp: $3.0\ \text{V}$