In a circuit with a 24V battery and two resistors of 8Ω and 4Ω in series, what is the voltage across the 4Ω resistor?

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Q: In a circuit with a 24V battery and two resistors of 8Ω and 4Ω in series, what is the voltage across the 4Ω resistor?

Solution: Using the voltage divider rule, V_R2 = (R2 / (R1 + R2)) * Vtotal = (4 / (8 + 4)) * 24 = 8V.

Steps: 8

Step 1: Identify the total voltage in the circuit, which is given as 24V from the battery.
Step 2: Identify the resistors in the circuit. We have R1 = 8Ω and R2 = 4Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: R_total = R1 + R2 = 8Ω + 4Ω = 12Ω.
Step 4: Use the voltage divider rule to find the voltage across the 4Ω resistor (R2). The formula is V_R2 = (R2 / R_total) * V_total.
Step 5: Substitute the values into the formula: V_R2 = (4Ω / 12Ω) * 24V.
Step 6: Simplify the fraction: 4 / 12 = 1 / 3.
Step 7: Multiply by the total voltage: V_R2 = (1 / 3) * 24V = 8V.
Step 8: Conclude that the voltage across the 4Ω resistor is 8V.