Scenario 1: A simple circuit contains a battery, a resistor, an ammeter, and a voltmeter. The ammeter reads 0.04 A and the voltmeter reads 6 V. What power is dissipated in the resistor R?
This equation does not need modification since the information in the scenario is given to solve for power.
You have been given two of the variables in the scenario, but they are not in the order of the equation. You must algebraically alter the equation to solve for V.
Scenario 1: A simple circuit contains a battery, a resistor, an ammeter, and a voltmeter. The ammeter reads 0.04 A and the voltmeter reads 6 V. What power is dissipated in the resistor R?
Scenario 2: A simple circuit contains a battery, a resistor, an ammeter, and a voltmeter. The ammeter reads 0.06 A. The power dissipated in the resistor R is 0.36 W. What is the voltage?
Scenario 3 Challenge Problem: A 6 V battery is connected to a 240 Ω resistor. What power is dissipated in the resistor?
Formula 1:
Scenario 3 Challenge Problem: A 6 V battery is connected to a 240 Ω resistor. What power is dissipated in the resistor?
Formula 2:
Start
Equation:
P = VI
Given
P = ?
V = 6 V
I = 0.04 A
Start
Equation:
P = VI
Given
P = 0.36 W
V = ?
I = 0.06 A
Start
Which two equations will you need to use? See hint for assistance.
Given
P = ?
V = 6 V
I = ?
R = 240 ohms
Start
Which two equations will you need to use? See hint for assistance.
Given
P = ?
V = 6 V
I = 0.025
R = 240 ohms
Answer:
P = (0.04 A) × (6 V)
P = 0.24 W
Answer:
0.36 W = (0.06 A) × (?)
0.36 W ÷ 0.06 A = V
V = 6 V
Answer:
I = V/R
I = 6/240
I = 0.025
Answer:
P = VI
P = (6) (0.025)
P = 0.15 W
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