In a circuit with a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor? (2022)
Practice Questions
1 question
Q1
In a circuit with a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor? (2022)
3V
6V
9V
12V
Total resistance R_total = 3Ω + 6Ω = 9Ω. Current I = V/R_total = 9V/9Ω = 1A. Voltage across 6Ω = I * R = 1A * 6Ω = 6V.
Questions & Step-by-step Solutions
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Q
Q: In a circuit with a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor? (2022)
Solution: Total resistance R_total = 3Ω + 6Ω = 9Ω. Current I = V/R_total = 9V/9Ω = 1A. Voltage across 6Ω = I * R = 1A * 6Ω = 6V.
Steps: 3
Step 1: Identify the total resistance in the circuit. Since the resistors are in series, add their resistances together: 3Ω + 6Ω = 9Ω.
Step 2: Use Ohm's Law to find the current in the circuit. Ohm's Law states that Current (I) = Voltage (V) / Resistance (R). Here, V = 9V and R_total = 9Ω. So, I = 9V / 9Ω = 1A.
Step 3: Calculate the voltage across the 6Ω resistor. Use the formula Voltage (V) = Current (I) * Resistance (R). Here, I = 1A and R = 6Ω. So, Voltage across 6Ω = 1A * 6Ω = 6V.
