In a circuit with a 9V battery and two resistors in series (R1 = 3Ω, R2 = 6Ω), w

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Question: In a circuit with a 9V battery and two resistors in series (R1 = 3Ω, R2 = 6Ω), what is the voltage across R2?

Options:

Correct Answer: 6V

Solution:

Using voltage division, V_R2 = (R2 / (R1 + R2)) * V_total = (6Ω / (3Ω + 6Ω)) * 9V = 6V.

In a circuit with a 9V battery and two resistors in series (R1 = 3Ω, R2 = 6Ω), what is the voltage across R2?

In a circuit with a 9V battery and two resistors in series (R1 = 3Ω, R2 = 6Ω), what is the voltage across R2?

Step 1: Identify the total voltage from the battery, which is 9V.
Step 2: Identify the values of the resistors: R1 = 3Ω and R2 = 6Ω.
Step 3: Calculate the total resistance in the circuit by adding R1 and R2: Total Resistance = R1 + R2 = 3Ω + 6Ω = 9Ω.
Step 4: Use the voltage division formula to find the voltage across R2: V_R2 = (R2 / Total Resistance) * V_total.
Step 5: Substitute the values into the formula: V_R2 = (6Ω / 9Ω) * 9V.
Step 6: Simplify the calculation: V_R2 = (6 / 9) * 9V = 6V.

Voltage Division – The principle that the voltage across a resistor in a series circuit is proportional to its resistance relative to the total resistance.