If a circuit has a 12V battery and two resistors of 3Ω and 6Ω in series, what is

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Question: If a circuit has a 12V battery and two resistors of 3Ω and 6Ω in series, what is the total current flowing through the circuit?

Options:

Correct Answer: 2A

Solution:

Total resistance R = R1 + R2 = 3Ω + 6Ω = 9Ω. Using Ohm\'s law, I = V/R = 12V / 9Ω = 1.33A.

If a circuit has a 12V battery and two resistors of 3Ω and 6Ω in series, what is the total current flowing through the circuit?

If a circuit has a 12V battery and two resistors of 3Ω and 6Ω in series, what is the total current flowing through the circuit?

Correct Answer: 1.33A

Step 1: Identify the voltage of the battery, which is 12V.
Step 2: Identify the resistors in the circuit, which are 3Ω and 6Ω.
Step 3: Calculate the total resistance by adding the two resistors together: 3Ω + 6Ω = 9Ω.
Step 4: Use Ohm's law to find the current. Ohm's law states that current (I) equals voltage (V) divided by resistance (R).
Step 5: Substitute the values into the formula: I = 12V / 9Ω.
Step 6: Calculate the current: 12V divided by 9Ω equals approximately 1.33A.

Ohm's Law – Ohm's Law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor.
Series Circuits – In a series circuit, the total resistance is the sum of the individual resistances, and the same current flows through all components.