1. Study the circuit shown and answer the following questions (the capacitor is completely discharged at t=0)   'R=0.8 M OHM' 'C=5uF'?

b) zero
c) battery voltage v₀ (not stated)
d) v = v₀[1–e^(–t/τ)] = v₀[1–e^(–1)]
v = 0.63v₀
Q = CV = 0.63v₀(5µ) = v₀(3.15) µC
e) Q = CV = v₀(5µ) = v₀(5) µC


voltage on a cap, charging
v = v₀[1–e^(–t/τ)]
v₀ is the battery voltage
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 = 0.63v₀
After 2τ, v = 0.86v₀
After 3τ, v = 0.95v₀
After 5τ, v = 0.993v₀...Show more