In a series circuit with a 9V battery and two resistors (3Ω and 6Ω), what is the

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

Options:

Correct Answer: 2A

Solution:

Total resistance R = 3Ω + 6Ω = 9Ω; I = V / R = 9V / 9Ω = 1A.

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

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

Step 1: Identify the components in the circuit. We have a 9V battery and two resistors: one with a resistance of 3Ω and another with a resistance of 6Ω.
Step 2: Calculate the total resistance in the circuit. Since this is a series circuit, we add the resistances together: 3Ω + 6Ω = 9Ω.
Step 3: Use Ohm's Law to find the current. Ohm's Law states that current (I) is equal to voltage (V) divided by resistance (R). Here, V = 9V and R = 9Ω.
Step 4: Substitute the values into the formula: I = 9V / 9Ω.
Step 5: Calculate the current: I = 1A.

Ohm's Law – The relationship between voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as I = V / R.
Series Circuit – A circuit configuration where components are connected end-to-end, resulting in the same current flowing through each component.
Total Resistance in Series – In a series circuit, the total resistance is the sum of the individual resistances.