Question

A(3.72×10 )−μF capacitor is connected to an AC source of EMF wit a frequency of (3.291×10²)Hz and a maximum EMF of (4.480×10

, V. What is the maximum current expressed in A and with three significant figures.

Answer 1

To determine the maximum current in an AC circuit with a given frequency and voltage, we must first calculate the capacitive reactance using the formula X_C = 1/(2πfC), and then apply Ohm's law for AC circuits, I_max = V_max / X_C. After performing the calculations and rounding, the answer will be expressed in amperes (A) with three significant figures.

Step-by-step explanation:

The question asks for the maximum current that flows through a capacitor connected to an alternating current (AC) source with a certain frequency and maximum voltage.

To find the maximum current, we first need to calculate the capacitive reactance (XC) using the formula XC = 1 / (ωC), where ω = 2πf is the angular frequency and C is the capacitance.

Then, we use Ohm's law for AC circuits to find the maximum current: Imax = Vmax / XC.

First, let's calculate the capacitive reactance:

The frequency is f = 3.291×102 Hz

The capacitance is C = 3.72×10−6 F

ω = 2πf = 2π×3.291×102

XC = 1 / (ωC)

Then, the AC source's maximum voltage (Vmax) is:

Vmax = 4.480×102 V

Now, we can calculate the maximum current Imax using:

Imax = Vmax / XC

After calculating XC and plugging in the values for Vmax, the maximum current can be determined, rounded to three significant figures, expressed in amperes (A).