In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms), what is the voltage drop across the 6 ohm resistor?
Practice Questions
1 question
Q1
In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms), what is the voltage drop across the 6 ohm resistor?
8 V
4 V
6 V
12 V
The total resistance is 9 ohms. The current is I = V/R = 12V / 9 ohms = 4/3 A. The voltage drop across the 6 ohm resistor is V = I * R = (4/3 A) * 6 ohms = 8 V.
Questions & Step-by-step Solutions
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Q
Q: In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms), what is the voltage drop across the 6 ohm resistor?
Solution: The total resistance is 9 ohms. The current is I = V/R = 12V / 9 ohms = 4/3 A. The voltage drop across the 6 ohm resistor is V = I * R = (4/3 A) * 6 ohms = 8 V.
Steps: 3
Step 1: Identify the total resistance in the circuit. You have two resistors in series: 3 ohms and 6 ohms. Add them together: 3 ohms + 6 ohms = 9 ohms.
Step 2: Use Ohm's Law to find the current in the circuit. Ohm's Law states that I = V / R. Here, V is the total voltage (12V) and R is the total resistance (9 ohms). So, I = 12V / 9 ohms = 4/3 A.
Step 3: Calculate the voltage drop across the 6 ohm resistor using Ohm's Law again. The formula is V = I * R. You already found I (4/3 A) and R for the 6 ohm resistor is 6 ohms. So, V = (4/3 A) * 6 ohms = 8 V.
