Using Kirchhoff's Voltage Law, if a loop in a circuit has a 12V battery and two resistors of 4Ω and 6Ω, what is the voltage across the 4Ω resistor?

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4.8V
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6V
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Q: Using Kirchhoff's Voltage Law, if a loop in a circuit has a 12V battery and two resistors of 4Ω and 6Ω, what is the voltage across the 4Ω resistor?

Solution: Using voltage division, V_R1 = (R1 / (R1 + R2)) * V_total = (4Ω / (4Ω + 6Ω)) * 12V = 4.8V.

Steps: 6

Step 1: Identify the total voltage in the circuit, which is given as 12V from the battery.
Step 2: Identify the resistors in the circuit. We have R1 = 4Ω and R2 = 6Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors: R_total = R1 + R2 = 4Ω + 6Ω = 10Ω.
Step 4: Use the voltage division formula to find the voltage across the 4Ω resistor (R1). The formula is V_R1 = (R1 / R_total) * V_total.
Step 5: Substitute the values into the formula: V_R1 = (4Ω / 10Ω) * 12V.
Step 6: Calculate the voltage across the 4Ω resistor: V_R1 = 0.4 * 12V = 4.8V.

Using Kirchhoff's Voltage Law, if a loop in a circuit has a 12V battery and two resistors of 4Ω and 6Ω, what is the voltage across the 4Ω resistor?

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