Question

The voltage across a 1μF capacitor is v= 30 sin 400t What is the sinusoidal expression for the current. Sketch the vand i curves.

Answer 1

The sinusoidal expression for the current in a circuit with a 1μF capacitor and voltage v(t) = 30 sin 400t is i(t) = (1μF) * (12000 cos 400t).

Step-by-step explanation:

To find the sinusoidal expression for the current in a circuit with a 1μF capacitor and voltage v(t) = 30 sin 400t, we need to use Ohm's Law and the relationship between voltage and current in a capacitor.

In a capacitor, the current is related to the rate of change of voltage concerning time. The current is given by the equation i(t) = C * dV(t)/dt, where C is the capacitance. Since the voltage across the capacitor in this case is v(t) = 30 sin 400t, we can find the derivative of this equation to find the current.

Taking the derivative of v(t) = 30 sin 400t gives us dv(t)/dt = 12000 cos 400t. Multiplying this by the capacitance, we get the expression for the current as i(t) = (1μF) * (12000 cos 400t).

Answer 2

a. The sinusoidal expression for the current is i = 0.012cos400t

b. Find the v and i curves in the attachment

Since the voltage across a 1μF capacitor is v= 30 sin 400t

a. What is the sinusoidal expression for the current

We know that the current across a capacitor is given by i = Cdv(t)/dt where C = capacitance = 1μF = 1 × 10⁻⁶ F

So, i = Cdv(t)/dt

i = 1 × 10⁻⁶ Fd30sin400t/dt

i = 1 × 10⁻⁶ F × 30dsin400t/dt

i = 1 × 10⁻⁶ F × 30 × 400cos400t

i = 1 × 10⁻⁶ F × 12000cos400t

i = 0.012cos400t

So, the sinusoidal expression for the current is i = 0.012cos400t

b. Sketch the v and i curves

Find the curves in the attachment