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.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach