Best Answer:
voltage on a cap, discharging

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

After 1τ, v = 0.37v₀

After 2τ, v = 0.14v₀

After 3τ, v = 0.05v₀

After 5τ, v = 0.807v₀

Energy in a Capacitor in Joules

E = ½CV² = ½QV = ½Q²/C

Q = CV

Q is charge in coulombs

C is capacitance in Farads

V is voltage in volts

E is energy in Joules

E = ½CV²

97.7m = ½C26.5²

C = 0.0977x2/26.5² = 0.000278 = 278 µF

What is the voltage at 1/2 energy?

E = ½CV₁²

E/2 = ½C(V₂)²

E =C(V₂)²

C(V₂)² = ½CV₁²

V₂² = ½V₁²

V₁²/V₂² = 2

V₁/V₂ = √2 = 1.414

so reducing the voltage by 1/√2 will reduce energy by 1/2

0.707 x 26.5 = 18.7 volts

v = v₀e^(–t/τ)

v/v₀ = e^(–t/τ) = 0.707

ln both sides

–t/τ = -0.347

τ = 3.8/0.347 = 10.96 sec

τ = RC = Rx278 µF

R = 10.96/278 µF = 0.0394 MΩ = 39.4 kΩ

but this is total, and 9.06 kΩ is already present

so answer = 39.4 kΩ – 9.06 kΩ = 30.4 kΩ

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