In a series circuit with a 12V battery and two resistors of 4Ω and 8Ω, what is the voltage across the 8Ω resistor? (2022)
Practice Questions
1 question
Q1
In a series circuit with a 12V battery and two resistors of 4Ω and 8Ω, what is the voltage across the 8Ω resistor? (2022)
4V
6V
8V
12V
Total resistance R_total = 4Ω + 8Ω = 12Ω. Current I = V/R_total = 12V/12Ω = 1A. Voltage across 8Ω = I * R = 1A * 8Ω = 8V.
Questions & Step-by-step Solutions
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Q
Q: In a series circuit with a 12V battery and two resistors of 4Ω and 8Ω, what is the voltage across the 8Ω resistor? (2022)
Solution: Total resistance R_total = 4Ω + 8Ω = 12Ω. Current I = V/R_total = 12V/12Ω = 1A. Voltage across 8Ω = I * R = 1A * 8Ω = 8V.
Steps: 3
Step 1: Identify the total resistance in the circuit. Add the resistance values of the two resistors: 4Ω + 8Ω = 12Ω.
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 (12Ω). So, I = 12V / 12Ω = 1A.
Step 3: Calculate the voltage across the 8Ω resistor using Ohm's Law again. The formula is V = I * R. Here, I is the current (1A) and R is the resistance of the 8Ω resistor. So, V = 1A * 8Ω = 8V.
