If $ C_1 = C_2 = 4.00 \, \mu F $ and $ C_4 = 8.00 \, \mu F $, what must the capacitance $ C_3 $ be if the network is to store $ 2.40 \times 10^{-3} \, J $ of electrical energy?

a) $ 2.00 \, \mu F $
b) $ 4.00 \, \mu F $
c) $ 6.00 \, \mu F $
d) $ 8.00 \, \mu F $

In order to determine the capacitance C₃ that will store 2.40 × 10⁻³ J of electrical energy, we need to use the formula for the energy stored in a capacitor:

U = ½C(V²)

Since the network is fully charged, the total energy stored will be the sum of the energy stored in each individual capacitor. We can use the formula to calculate the energy stored in each capacitor:

U = ½C₁(V₁²)

U = ½C₂(V₂²)

U = ½C₃(V₃²)

Since C₁ = C₂ = 4.00 μF and C₄ = 8.00 μF, and the energy stored is given as 2.40 × 10⁻³ J, we can substitute those values into the equations:

2.40 × 10⁻³ J = ½(4.00 μF)(V₁²)

2.40 × 10⁻³ J = ½(4.00 μF)(V₂²)

2.40 × 10⁻³ J = ½(C₃)(V₃²)

Since the potential voltage across the capacitors is the same, we can solve for V₁ and V₂ and substitute their values into the equation:

V₁ = V₂ = V

2.40 × 10⁻³ J = ½(4.00 μF)(V²)

V² = (2.40 × 10⁻³ J)/(2.00 μF)

V² = (1.20 × 10⁻³ J/μF)

Now, we can solve for V and substitute it into the equation for C₃:

2.40 × 10⁻³ J = ½(C₃)(V²)

2.40 × 10⁻³ J = ½(C₃)(1.20 × 10⁻³ J/μF)

2.00 = C₃

Therefore, the capacitance C₃ must be 2.00 μF in order for the network to store 2.40 × 10⁻³ J of electrical energy.

If \( C_1 = C_2 = 4.00 \, \mu F \) and \( C_4 = 8.00 \, \mu F \), what must the capacitance \( C_3 \) be if the network is to store \( 2.40 \times 10^{-3} \, J \) of electrical energy?a) \( 2.00 \, \mu F \) b) \( 4.00 \, \mu F \) c) \( 6.00 \, \mu F \) d) \( 8.00 \, \mu F \)

Answer :

Final answer:

Explanation:

Other Questions

If \( C_1 = C_2 = 4.00 \, \mu F \) and \( C_4 = 8.00 \, \mu F \), what must the capacitance \( C_3 \) be if the network is to store \( 2.40 \times 10^{-3} \, J \) of electrical energy?

a) \( 2.00 \, \mu F \)
b) \( 4.00 \, \mu F \)
c) \( 6.00 \, \mu F \)
d) \( 8.00 \, \mu F \)