A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?
Practice Questions
1 question
Q1
A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?
3 V
6 V
9 V
4.5 V
Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.
Questions & Step-by-step Solutions
1 item
Q
Q: A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?
Solution: Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.
Steps: 4
Step 1: Identify the components in the circuit. We have a 9V battery and two resistors: one is 3 ohms and the other is 6 ohms.
Step 2: Calculate the total resistance in the circuit. Since the resistors are in series, add their resistances together: 3 ohms + 6 ohms = 9 ohms.
Step 3: Use Ohm's Law to find the current in the circuit. Ohm's Law states that current (I) equals voltage (V) divided by resistance (R). Here, V is 9V and R_total is 9 ohms. So, I = 9V / 9 ohms = 1 A.
Step 4: Now, calculate the voltage drop across the 6 ohm resistor. Use the formula: Voltage drop = Current (I) * Resistance (R). Here, I is 1 A and R is 6 ohms. So, Voltage drop = 1 A * 6 ohms = 6 V.
