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A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to discharge completely. Part A How much energy is dissipated by the 55Ω resistor?

Respuesta :

onyebuchinnaji

The energy dissipated by the 55 Ω resistor is 5.07 x 10⁻⁴ J.

The given parameters;

  • capacitance, C = 0.75 μF
  • voltage, V = 70 V
  • resistance, R₁ = 55Ω and R₁ = 140 Ω

The total energy stored in the capacitor is calculated as follows;

E = ¹/₂CV²

E = ¹/₂ x 0.75 x 10⁻⁶ x 70²

E = 0.0018 J

The the equivalent resistance is calculated as follows;

R = 55Ω + 140Ω = 195 Ω

The current flowing in the circuit is calculated as follows;

[tex]V= IR\\\\I = \frac{V}R} = \frac{70}{195} = 0.359 \ A[/tex]

The time taken for the energy to be dissipated is calculated as follows;

[tex]E = IVt\\\\t = \frac{E}{IV} = \frac{0.0018}{0.359 \times 70} \\\\t = 7.16 \times 10^{-5} \ s[/tex]

The energy dissipated by the 55 Ω resistor is calculated as follows;

[tex]E_1 = I^2R_1 t\\\\E_1 =( 0.359)^2 \times (55)\times (7.16\times 10^{-5})\\\\E_1 = 5.07 \times 10^{-4} \ J[/tex]

Thus, the energy dissipated by the 55 Ω resistor is 5.07 x 10⁻⁴ J.

Learn more here:https://brainly.com/question/15319056

hamzaahmeds

This question involves the concepts of energy of capacitor, Ohm's Law, Joule's Law, Energy Dissipated, and Equivalent Resistance.

The Energy dissipated by the 55Ω resistor is "5.1 x 10⁻⁴ J".

First, we calculate the total energy of capacitor:

[tex]E = \frac{1}{2}CV^2\\\\[/tex]

where,

E = energy of capacitor = ?

C = Capacitance = 0.75 μF = 7.5 x 10⁻⁷ F

V = Voltage = 70 V

Therefore,

[tex]E=\frac{1}{2}(7.5\ x\ 10^{7}\ F)(70\ V)^2[/tex]

E = 0.0018 J

The equivalent resistance in series combination can be calculated as follows:

R = R₁ + R₂ = 55 Ω + 140 Ω

R = 195 Ω

From Ohm's Law:

[tex]V=IR\\\\I=\frac{V}{R}\\\\I=\frac{70\ V}{195\ \Omega}[/tex]

I = 0.36 A

Now, from Joule's Law:

[tex]E = I^2Rt\\\\t=\frac{E}{I^2R}\\\\t=\frac{0.0018\ J}{(0.36\ A)^2(195\ \Omega)}\\\\[/tex]

t = 7.16 x 10⁻⁵ s

Now, the energy dissipated in R₁ (55 Ω resistor) is:

Ε₁ = I²R₁t

E₁ = (0.36 A)²(55 Ω)(7.16 x 10⁻⁵ s)

E₁ = 5.1 x 10⁻⁴ J = 0.51 mJ

Learn more about Ohm's Law here:

brainly.com/question/17286882?referrer=searchResults

The attached picture shows Ohm's Law.

Ver imagen hamzaahmeds

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