Respuesta :
Answer:
(a) RC time constant of the circuit is 6.9 × 10⁻⁶ ms
(b) The potential drop across the capacitor 1.00 s after the switch is closed is 0 V
Explanation:
The given parameters are
The length of the graphite pencil, L = 5.00 cm
The diameter of the graphite, D = 0.05 mm
The resistivity of the graphite, ρ = 1.38 × 10⁻⁵ Ω/m
The capacitance of the capacitor, C = 10.0 mF
The voltage of the power source, V = 0.50-V
(a) The RC time constant of the circuit, τ, is given as follows;
τ = R × C
Where;
R = The resistance of the graphite = L × ρ
C = The capacitance of the capacitor
∴ R = 5.00 cm × 1.38 × 10⁻⁵ Ω/m = 6.9 × 10⁻⁷ Ω
RC time constant of the circuit, τ = 6.9 × 10⁻⁷ Ω × 10.0 mF = 6.9 × 10⁻⁶ ms
RC time constant of the circuit, τ = 6.9 × 10⁻⁶ ms
(b) The potential drop after t = 1.00 s is given as follows;
[tex]i = \dfrac{V}{R} \cdot e^{-\dfrac{t}{R\cdot C} }[/tex]
Where;
I = The current in the circuit
V = The voltage in the circuit = 0.50 V
R = resistance in the circuit = 6.9 × 10⁻⁷ Ω
C = The series capacitance = 10.0 mF
t = The time taken = 1.00 s
Plugging in the variable values, gives;
[tex]I = \dfrac{0.5}{6.9 \times 10^{-7}} \cdot e^{-\dfrac{1.00}{6.9 \times 10^{-7}\times 10.0 \ mF} } = 0[/tex]
V(1) = I·R = 0 × R = 0
The potential drop across the capacitor 1.00 s after the switch is closed, V(1) = 0 V
