Answer:
A. 0.199 J
B. 0.0663 C
C = 0.0221 F
D. 12.68 ohms
Step-by-step explanation:
From the question:
time duration, t = 0.28 seconds
Average power, P = 0.71 W
Average voltage, V = 3 V
A) Energy is given as:
E = P * t
=> E = 0.71 * 0.28 = 0.199 J
B) Electrical energy is also given as:
E = qV
where q = charge
=> q = E / V
∴ q = 0.199 / 3 = 0.0663 C
C) Capacitance is given as charge over voltage:
C = q / V
=> C = 0.0663 / 3 = 0.0221 F
D) Electrical power, P, can also be given as:
P =
![V^2 / R](https://img.qammunity.org/2021/formulas/physics/college/b82k22mpup8wdt9gbw1s21ht5cbiaei9q2.png)
where R = resistance
=> R =
![V^2 / P](https://img.qammunity.org/2021/formulas/physics/college/hn7a7bzac9bajiuh054q6tqvdzpzj4cvq5.png)
R =
![3^2 / 0.71 = 9 / 0.71 = 12.68 ohms](https://img.qammunity.org/2021/formulas/physics/college/bkmh9hl3czd8p4s8ojdogo6t5ukzo9ucr6.png)