In a circuit with a 24V battery and two resistors in series (R1 = 6Ω, R2 = 12Ω), what is the voltage drop across R2?

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Q: In a circuit with a 24V battery and two resistors in series (R1 = 6Ω, R2 = 12Ω), what is the voltage drop across R2?

Solution: Voltage drop across R2 = V_R2 = (R2 / (R1 + R2)) * V_total = (12Ω / (6Ω + 12Ω)) * 24V = 12V.

Steps: 8

Step 1: Identify the total voltage from the battery, which is 24V.
Step 2: Identify the values of the resistors: R1 = 6Ω and R2 = 12Ω.
Step 3: Calculate the total resistance in the circuit by adding R1 and R2: Total Resistance = R1 + R2 = 6Ω + 12Ω = 18Ω.
Step 4: Use the formula to find the voltage drop across R2: Voltage drop across R2 = (R2 / Total Resistance) * V_total.
Step 5: Substitute the values into the formula: Voltage drop across R2 = (12Ω / 18Ω) * 24V.
Step 6: Simplify the fraction: 12Ω / 18Ω = 2/3.
Step 7: Multiply: (2/3) * 24V = 16V.
Step 8: Therefore, the voltage drop across R2 is 16V.