A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series

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

Options:

Correct Answer: 6 V

Solution:

Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.

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

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

Correct Answer: 6 V

Step 1: Identify the components in the circuit. We have a 9V battery and two resistors: one is 3 ohms and the other is 6 ohms.
Step 2: Calculate the total resistance in the circuit. Since the resistors are in series, add their resistances together: 3 ohms + 6 ohms = 9 ohms.
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 is 9V and R_total is 9 ohms. So, I = 9V / 9 ohms = 1 A.
Step 4: Now, calculate the voltage drop across the 6 ohm resistor. Use the formula: Voltage drop = Current (I) * Resistance (R). Here, I is 1 A and R is 6 ohms. So, Voltage drop = 1 A * 6 ohms = 6 V.

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.