If a circuit has a 15V battery and two resistors of 5Ω and 10Ω in series, what is the power dissipated by the 10Ω resistor?
Practice Questions
1 question
Q1
If a circuit has a 15V battery and two resistors of 5Ω and 10Ω in series, what is the power dissipated by the 10Ω resistor?
1.5W
2.5W
3.75W
5W
Total resistance R = 5Ω + 10Ω = 15Ω. Current I = V/R = 15V/15Ω = 1A. Power P = I^2 * R = 1^2 * 10 = 10W.
Questions & Step-by-step Solutions
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Q
Q: If a circuit has a 15V battery and two resistors of 5Ω and 10Ω in series, what is the power dissipated by the 10Ω resistor?
Solution: Total resistance R = 5Ω + 10Ω = 15Ω. Current I = V/R = 15V/15Ω = 1A. Power P = I^2 * R = 1^2 * 10 = 10W.
Steps: 3
Step 1: Identify the total resistance in the circuit. Since the resistors are in series, add their resistances together: 5Ω + 10Ω = 15Ω.
Step 2: Use Ohm's Law to find the current in the circuit. The formula is I = V/R. Here, V is the voltage of the battery (15V) and R is the total resistance (15Ω). So, I = 15V / 15Ω = 1A.
Step 3: Calculate the power dissipated by the 10Ω resistor. Use the formula P = I^2 * R. We already found I = 1A, and R for the 10Ω resistor is 10Ω. So, P = (1A)^2 * 10Ω = 1 * 10 = 10W.
