In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is

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Question: In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is the voltage drop across the 3Ω resistor?

Options:

Correct Answer: 6V

Solution:

Using the voltage divider rule, V3 = (R3 / (R2 + R3)) * Vtotal = (3 / (2 + 3)) * 10 = 6V.

In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is the voltage drop across the 3Ω resistor?

In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is the voltage drop across the 3Ω resistor?

Step 1: Identify the total voltage in the circuit, which is given as 10V from the battery.
Step 2: Identify the resistors in the circuit. We have R2 = 2Ω and R3 = 3Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: Total Resistance = R2 + R3 = 2Ω + 3Ω = 5Ω.
Step 4: Use the voltage divider rule to find the voltage drop across the 3Ω resistor. The formula is V3 = (R3 / (R2 + R3)) * Vtotal.
Step 5: Substitute the values into the formula: V3 = (3 / (2 + 3)) * 10.
Step 6: Simplify the equation: V3 = (3 / 5) * 10.
Step 7: Calculate the result: V3 = 6V.

Voltage Divider Rule – The voltage divider rule is used to determine the voltage drop across a resistor in a series circuit based on the ratio of its resistance to the total resistance.
Series Circuit – In a series circuit, the total voltage is divided among the resistors, and the same current flows through each resistor.