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Makovka662 [10]
3 years ago
12

Two identical capacitors are connected parallel. Initially they are charged to a potential V₀ and each acquired a charge Q₀. The

battery is then disconnected, and the gap between the plates of one capacitor is filled with a dielectric κ.(a) What is the new potential difference V across the capacitors. possible asnwers: V=(V₀)^2/[kQ₀+V₀), V=V₀/2k, V=V₀/2, V=kQ₀/V₀, V=2V₀/[k+1](b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Physics
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

ΔV ’= 2Δv / (1 + k) ,            Q’= Q₀ 1,772

Explanation:

When the capacitors are connected in series the capacitance is added

      C_eq = C₁ + C₂2

     C_eq = 2 C₀

     ΔV = Q₀ / C_eq

     ΔV = Q₀ / 2 C₀

In this case, a dielectric is introduced to one of the capacitors, so its capacity changes.

         C₂’= k C₂

Since the two capacitors have the same initial value let's call Co

          C_eq’= C₀ (1 + k)

The set charge is

           C_eq ’= Q₀ / ΔV’

            ΔV ’= Q₀ / Ceq’

            ΔV ’= Q₀ / C₀ (1 + k)

The relationship between these voltages is

            ΔV'/ ΔV = 2 / (1 + k)

            ΔV ’= 2Δv / (1 + k)

The initial charge of the capacitor is

              Q₀ = C₀ ΔV

               

The charge after introducing the dielectric is

             Q’= k C₀ ΔV’

             Q ’= k C₀ 2 ΔV / (1 + k)

             Q’= Q₀ 2k / (1 + k)

             Q’= Q₀  2 7.8 /(1+7.8)

              Q’= Q₀ 1,772

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Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

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The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

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When connected in series, the power supplied is given as

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R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

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