In a circuit with a 24V battery and three resistors of 4Ω, 4Ω, and 8Ω in series, what is the voltage drop across the 8Ω resistor?
Practice Questions
1 question
Q1
In a circuit with a 24V battery and three resistors of 4Ω, 4Ω, and 8Ω in series, what is the voltage drop across the 8Ω resistor?
8V
12V
16V
24V
Total resistance R = 4Ω + 4Ω + 8Ω = 16Ω. Current I = V/R = 24V/16Ω = 1.5A. Voltage drop across 8Ω = I * R = 1.5A * 8Ω = 12V.
Questions & Step-by-step Solutions
1 item
Q
Q: In a circuit with a 24V battery and three resistors of 4Ω, 4Ω, and 8Ω in series, what is the voltage drop across the 8Ω resistor?
Solution: Total resistance R = 4Ω + 4Ω + 8Ω = 16Ω. Current I = V/R = 24V/16Ω = 1.5A. Voltage drop across 8Ω = I * R = 1.5A * 8Ω = 12V.
Steps: 6
Step 1: Identify the total resistance in the circuit. Add the resistances of all three resistors: 4Ω + 4Ω + 8Ω.
Step 2: Calculate the total resistance: 4Ω + 4Ω + 8Ω = 16Ω.
Step 3: Use Ohm's Law to find the current in the circuit. The formula is I = V/R, where V is the voltage of the battery (24V) and R is the total resistance (16Ω).
Step 4: Calculate the current: I = 24V / 16Ω = 1.5A.
Step 5: To find the voltage drop across the 8Ω resistor, use the formula: Voltage drop = Current (I) * Resistance (R).
Step 6: Calculate the voltage drop across the 8Ω resistor: Voltage drop = 1.5A * 8Ω = 12V.
