In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series

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Question: In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?

Options:

Correct Answer: 6V

Solution:

Total resistance R_total = 4 + 8 = 12 ohms. Current I = V/R_total = 12V/12 ohms = 1A. Voltage drop across 8 ohm resistor = I * R = 1A * 8 ohms = 8V.

In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?

In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?

Correct Answer: 8V

Step 1: Identify the total resistance in the circuit. Add the resistance values of the two resistors: 4 ohms + 8 ohms = 12 ohms.
Step 2: Use Ohm's Law to find the current in the circuit. The formula is I = V / R_total. Here, V is the battery voltage (12V) and R_total is the total resistance (12 ohms). So, I = 12V / 12 ohms = 1A.
Step 3: Calculate the voltage drop across the 8 ohm resistor. Use the formula V = I * R, where I is the current (1A) and R is the resistance of the 8 ohm resistor. So, V = 1A * 8 ohms = 8V.

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.