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