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?

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

The given parameters;

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;

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

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

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

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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:

where,

E = energy of capacitor = ?

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

V = Voltage = 70 V

Therefore,

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:

I = 0.36 A

Now, from Joule's Law:

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:

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The attached picture shows Ohm's Law.

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explanation: