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

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

Options:

Correct Answer: 2A

Solution:

Total resistance R_total = R1 + R2 = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V/9Ω = 1A.

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

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

Step 1: Identify the voltage of the battery. In this case, it is 9V.
Step 2: Identify the values of the resistors. R1 is 3Ω and R2 is 6Ω.
Step 3: Calculate the total resistance in the circuit. Add R1 and R2 together: R_total = R1 + R2 = 3Ω + 6Ω.
Step 4: Perform the addition: R_total = 9Ω.
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 the formula: I = 9V / 9Ω.
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.