A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of
1 answer:
You might be interested in
Given Information:
Potential difference = V = 100 V
Capacitance C₁ = 10 mF
Capacitance C₂ = 5 mF
Capacitance C₃ = 4 mF
Required Information:
a. Charge q₃
b. Potential difference V₃
c. Stored energy U₃
d. Charge q ₁
e. Potential difference V₁
f. Stored energy U₁
g. Charge q ₂
h. Potential difference V₂
i. Stored energy U₂
Answer:
a. Charge q₃ = 0.4 C
b. Potential difference V₃ = 100 V
c. Stored energy U₃ = 20 J
d. Charge q ₁ = 0.33 C
e. Potential difference V₁ = 33 V
f. Stored energy U₁ = 5.445 J
g. Charge q ₂ = 0.33 C
h. Potential difference V₂ = 66 V
i. Stored energy U₂ = 10.89 J
Explanation:
Please refer to the circuit attached in the diagram
a. Charge q₃
As we know charge in a capacitor is given by
q₃ = C₃V₃
q₃ = 4x10⁻³*100
q₃ = 0.4 C
b. Potential difference V₃
The potential difference V₃ is same as V
V₃ = 100 V
c. Stored energy U₃
Energy stored in a capacitor is given by
U₃ = ½C₃V₃²
U₃ = ½*4x10⁻³*100²
U₃ = 20 J
d. Charge q ₁
Since capacitor C₁ and C₂ are in series their equivalent capacitance is
Ceq = C₁*C₂/C₁ + C₂
Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³
Ceq = 3.33x10⁻³ F
q ₁ = Ceq*V
q ₁ = 3.33x10⁻³*100
q ₁ = 0.33 C
e. Potential difference V₁
V₁ = q ₁/C₁
V₁ = 0.33/10x10⁻³
V₁ = 33 V
f. Stored energy U₁
U₁ = ½C₁V₁²
U₁ = ½*10x10⁻³*(33)²
U₁ = 5.445 J
g. Charge q ₂
q₂ = Ceq*V
q₂ = 3.33x10⁻³*100
q₂ = 0.33 C
h. Potential difference V₂
V₂ = q ₂/C₂
V₂ = 0.33/5x10⁻³
V₂ = 66 V
i. Stored energy U₂
U₂ = ½C₂V₂²
U₂ = ½*5x10⁻³*(66)²
U₂ = 10.89 J
