In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is

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

Options:

Correct Answer: 6V

Solution:

Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.

In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?

In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?

Step 1: Identify the total resistance in the circuit. Add the resistance values of the two resistors: 4Ω + 8Ω = 12Ω.
Step 2: Use Ohm's Law to find the current in the circuit. The formula is I = V / R, where V is the voltage of the battery (12V) and R is the total resistance (12Ω). So, I = 12V / 12Ω = 1A.
Step 3: Calculate the voltage drop across the 8Ω resistor using Ohm's Law again. The formula is Voltage drop = I * R, where I is the current (1A) and R is the resistance of the 8Ω resistor. So, Voltage drop = 1A * 8Ω = 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.
Voltage Division – In a series circuit, the voltage drop across a resistor is proportional to its resistance relative to the total resistance.