In a circuit with two resistors in series, if the total voltage across the circuit is 12V and the resistors are 4Ω and 6Ω, what is the voltage across the 4Ω resistor?
Practice Questions
1 question
Q1
In a circuit with two resistors in series, if the total voltage across the circuit is 12V and the resistors are 4Ω and 6Ω, what is the voltage across the 4Ω resistor?
4V
6V
8V
12V
Using the voltage divider rule, V1 = (R1 / (R1 + R2)) * Vtotal = (4 / (4 + 6)) * 12 = 4.8V.
Questions & Step-by-step Solutions
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Q
Q: In a circuit with two resistors in series, if the total voltage across the circuit is 12V and the resistors are 4Ω and 6Ω, what is the voltage across the 4Ω resistor?
Solution: Using the voltage divider rule, V1 = (R1 / (R1 + R2)) * Vtotal = (4 / (4 + 6)) * 12 = 4.8V.
Steps: 8
Step 1: Identify the total voltage in the circuit, which is given as 12V.
Step 2: Identify the resistors in the circuit. We have R1 = 4Ω and R2 = 6Ω.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: Rtotal = R1 + R2 = 4Ω + 6Ω = 10Ω.
Step 4: Use the voltage divider rule to find the voltage across the 4Ω resistor (V1). The formula is V1 = (R1 / Rtotal) * Vtotal.
Step 5: Substitute the values into the formula: V1 = (4Ω / 10Ω) * 12V.
Step 6: Calculate the fraction: 4 / 10 = 0.4.
Step 7: Multiply 0.4 by 12V to find V1: V1 = 0.4 * 12V = 4.8V.
Step 8: Conclude that the voltage across the 4Ω resistor is 4.8V.
