In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω),

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Question: In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the current flowing through the circuit?

Options:

Correct Answer: 1A

Solution:

Total resistance R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. Current I = V/R_total = 12V/12Ω = 1A.

In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the current flowing through the circuit?

In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the current flowing through the circuit?

Correct Answer: 1A

Step 1: Identify the values given in the problem. We have a 12V battery and two resistors: R1 = 4Ω and R2 = 8Ω.
Step 2: Calculate the total resistance in the circuit. Since the resistors are in series, we add them together: R_total = R1 + R2.
Step 3: Substitute the values of the resistors into the equation: R_total = 4Ω + 8Ω.
Step 4: Perform the addition: R_total = 12Ω.
Step 5: Use Ohm's Law to find the current. Ohm's Law states that Current (I) = Voltage (V) / Resistance (R).
Step 6: Substitute the values into Ohm's Law: I = 12V / 12Ω.
Step 7: Perform the division: I = 1A.
Step 8: Conclude that the current flowing through the circuit is 1A.

Ohm's Law – The relationship between voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = I * R.
Series Circuits – In a series circuit, the total resistance is the sum of individual resistances, and the same current flows through all components.