Question

A 1.50-mF capacitor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored in the capacitor is zero at t = 0, determine the magnitude of the current in the wires at t = 1/171s.

Answer 1

Answer:

96.10 A

Explanation:

The capacitor is connected to a voltage source defined by the expression,

V(t) = √2 × Vrms×sin2πft

Where V(t)=instantaneous voltage, Vrms= root mean square voltage value, f= frequency, t= time π=pi.

Vrms= 120 V, f= 60.0Hz, t= 1/171s, π= 3.143.

∴ V(t) = √2 × 120 × sin(2×3.143×60×t)

∴V(t) = 169.71 × sin(377.16t)............(1)

using capacitor's equation,

I(t)=C(dv/dt)...................(2)

Note: the differentiation of sin∅ = cos∅

∴ dv/dt=169.71×377.16×cos(377.16t)

  dv/dt = 64109.65× cos(377.16t)

  At time t=1/171s.

 ∴dv/dt=64109.65 × cos(377.16×1/171)

   dv/dt = 64109.65 ×cos(2.21)

   dv/dt = 64109.65 × 0.999

   dv/dt = 64045.54.

    I(t) = C(dv/dt)

   Where C= I.5 mF = 0.0015 F

   ∴ I(t) = 0.0015 × 64045.54

      I(t) = 96.07

      I(t) ≈ 96.10 A