Question 6 of 10
1 answer:
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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
To solve this problem we will apply Newton's second law and the principle of balancing Forces on the rope. Newton's second law allows us to define the weight of the mass, through the function
Here,
m = mass
a = g = Gravitational acceleration
Replacing we have that the weight is
Since the rope is taut and does not break, the net force on the rope will be zero.
Therefore the tensile force in the rope is 98N