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 drop across the 4Ω resistor?

Practice Questions

1 question

4.8V
8V
6V
3.2V

Questions & Step-by-step Solutions

1 item

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 drop across the 4Ω resistor?

Solution: Using voltage division, V_R1 = V_total * (R1 / (R1 + R2)) = 12V * (4Ω / (4Ω + 6Ω)) = 12V * (4/10) = 4.8V.

Steps: 7

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 drop across the 4Ω resistor (R1). The formula is V_R1 = V_total * (R1 / R_total).
Step 5: Substitute the values into the formula: V_R1 = 12V * (4Ω / 10Ω).
Step 6: Simplify the fraction: 4Ω / 10Ω = 0.4.
Step 7: Multiply to find the voltage drop: V_R1 = 12V * 0.4 = 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 drop across the 4Ω resistor?

Practice Questions

Questions & Step-by-step Solutions

Related Questions

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?