In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?
Practice Questions
1 question
Q1
In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?
4V
6V
8V
12V
Total resistance R_total = 4 + 8 = 12 ohms. Current I = V/R_total = 12V/12 ohms = 1A. Voltage drop across 8 ohm resistor = I * R = 1A * 8 ohms = 8V.
Questions & Step-by-step Solutions
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Q
Q: In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?
Solution: Total resistance R_total = 4 + 8 = 12 ohms. Current I = V/R_total = 12V/12 ohms = 1A. Voltage drop across 8 ohm resistor = I * R = 1A * 8 ohms = 8V.
Steps: 3
Step 1: Identify the total resistance in the circuit. Add the resistance values of the two resistors: 4 ohms + 8 ohms = 12 ohms.
Step 2: Use Ohm's Law to find the current in the circuit. The formula is I = V / R_total. Here, V is the battery voltage (12V) and R_total is the total resistance (12 ohms). So, I = 12V / 12 ohms = 1A.
Step 3: Calculate the voltage drop across the 8 ohm resistor. Use the formula V = I * R, where I is the current (1A) and R is the resistance of the 8 ohm resistor. So, V = 1A * 8 ohms = 8V.
