If two resistors of 4 ohms and 6 ohms are connected in series, what is the voltage drop across the 4 ohm resistor when a current of 2 A flows through the circuit?
Practice Questions
1 question
Q1
If two resistors of 4 ohms and 6 ohms are connected in series, what is the voltage drop across the 4 ohm resistor when a current of 2 A flows through the circuit?
8 V
4 V
2 V
6 V
Using Ohm's law, V = I * R, the voltage drop across the 4 ohm resistor is V = 2 A * 4 ohms = 8 V.
Questions & Step-by-step Solutions
1 item
Q
Q: If two resistors of 4 ohms and 6 ohms are connected in series, what is the voltage drop across the 4 ohm resistor when a current of 2 A flows through the circuit?
Solution: Using Ohm's law, V = I * R, the voltage drop across the 4 ohm resistor is V = 2 A * 4 ohms = 8 V.
Steps: 7
Step 1: Identify the values given in the problem. We have a 4 ohm resistor and a 6 ohm resistor connected in series, and a current of 2 A flowing through the circuit.
Step 2: Recall Ohm's law, which states that Voltage (V) = Current (I) * Resistance (R).
Step 3: For the 4 ohm resistor, we need to find the voltage drop across it using the current flowing through the circuit.
Step 4: Substitute the values into Ohm's law. Here, I = 2 A and R = 4 ohms.
Step 5: Calculate the voltage drop: V = 2 A * 4 ohms.
Step 6: Perform the multiplication: 2 * 4 = 8.
Step 7: Conclude that the voltage drop across the 4 ohm resistor is 8 V.
