A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What i

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Question: A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?

Options:

Correct Answer: 6V

Solution:

The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.

A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?

A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?

Correct Answer: 6V

Step 1: Identify the components in the circuit. We have a 9V battery and two resistors: one is 3Ω and the other is 6Ω.
Step 2: Calculate the total resistance in the circuit. Since the resistors are in series, add their resistances: 3Ω + 6Ω = 9Ω.
Step 3: Use Ohm's Law to find the current in the circuit. Ohm's Law states that current (I) equals voltage (V) divided by resistance (R). Here, V = 9V and R_total = 9Ω, so I = 9V / 9Ω = 1A.
Step 4: Calculate the voltage across the 6Ω resistor. Use the formula V = I * R, where I is the current we just calculated (1A) and R is the resistance of the 6Ω resistor. So, V = 1A * 6Ω = 6V.

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.