In a circuit with a 15V battery and two resistors in series (5Ω and 10Ω), what i

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Question: In a circuit with a 15V battery and two resistors in series (5Ω and 10Ω), what is the voltage drop across the 10Ω resistor?

Options:

Correct Answer: 10V

Solution:

Using the voltage divider rule, V2 = (R2 / (R1 + R2)) * Vtotal = (10 / (5 + 10)) * 15 = 10V.

In a circuit with a 15V battery and two resistors in series (5Ω and 10Ω), what is the voltage drop across the 10Ω resistor?

In a circuit with a 15V battery and two resistors in series (5Ω and 10Ω), what is the voltage drop across the 10Ω resistor?

Step 1: Identify the total voltage from the battery, which is 15V.
Step 2: Identify the resistors in the circuit. We have R1 = 5Ω and R2 = 10Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: Rtotal = R1 + R2 = 5Ω + 10Ω = 15Ω.
Step 4: Use the voltage divider rule to find the voltage drop across the 10Ω resistor (R2). The formula is V2 = (R2 / Rtotal) * Vtotal.
Step 5: Substitute the values into the formula: V2 = (10Ω / 15Ω) * 15V.
Step 6: Simplify the calculation: V2 = (10 / 15) * 15 = 10V.
Step 7: Conclude that the voltage drop across the 10Ω resistor is 10V.

Voltage Divider Rule – The voltage divider rule is used to determine the voltage drop across a resistor in a series circuit based on the resistance values and the total voltage.
Series Circuit – In a series circuit, the total resistance is the sum of individual resistances, and the same current flows through all components.