In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is

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Question: In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is the voltage drop across the 6Ω resistor?

Options:

Correct Answer: 6V

Solution:

Voltage drop across R2 = (R2 / (R1 + R2)) * Vtotal = (6 / (3 + 6)) * 10 = 6.67V.

In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is the voltage drop across the 6Ω resistor?

In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is the voltage drop across the 6Ω resistor?

Correct Answer: 6.67V

Step 1: Identify the total voltage from the battery, which is 10V.
Step 2: Identify the resistors in the circuit. We have R1 = 3Ω and R2 = 6Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: Rtotal = R1 + R2 = 3Ω + 6Ω = 9Ω.
Step 4: Use the formula to find the voltage drop across the 6Ω resistor (R2). The formula is: Voltage drop across R2 = (R2 / Rtotal) * Vtotal.
Step 5: Substitute the values into the formula: Voltage drop across R2 = (6Ω / 9Ω) * 10V.
Step 6: Calculate the fraction: 6 / 9 = 0.6667.
Step 7: Multiply this fraction by the total voltage: 0.6667 * 10V = 6.67V.
Step 8: Conclude that the voltage drop across the 6Ω resistor is 6.67V.

Voltage Division – The principle that the voltage drop across a resistor in a series circuit is proportional to its resistance relative to the total resistance.
Ohm's Law – The relationship between voltage, current, and resistance in an electrical circuit, expressed as V = IR.