Math Snap

STEP 1

1. The capacitor draws a current of 150 mA when connected to a voltage source.
2. The voltage source is $100 \, \mathrm{V}$ at a frequency of $400 \, \mathrm{Hz}$.
3. We are trying to find the capacitance of the capacitor.
4. The relationship between current, voltage, frequency, and capacitance in an AC circuit is given by the formula $I = V \cdot 2\pi f C$.

STEP 2

1. Convert all units to standard SI units.
2. Use the formula for current in an AC circuit to express capacitance.
3. Solve for capacitance.

STEP 3

Convert all units to standard SI units.
- Current: $150 \, \mathrm{mA} = 0.150 \, \mathrm{A}$
- Voltage: $100 \, \mathrm{V}$ (already in SI units)
- Frequency: $400 \, \mathrm{Hz}$ (already in SI units)

STEP 4

Use the formula for current in an AC circuit to express capacitance.
The formula relating current $I$, voltage $V$, frequency $f$, and capacitance $C$ is:
$$I = V \cdot 2\pi f C$$ Rearrange the formula to solve for $C$:
$$C = \frac{I}{V \cdot 2\pi f}$$

Substitute the known values into the formula and solve for $C$.
$$C = \frac{0.150}{100 \cdot 2\pi \cdot 400}$$ Calculate the denominator:
$$2\pi \cdot 400 = 800\pi$$ $$100 \cdot 800\pi = 80000\pi$$ Now calculate $C$:
$$C = \frac{0.150}{80000\pi}$$ $$C \approx \frac{0.150}{251327.4}$$ $$C \approx 0.000000597 \, \mathrm{F}$$ Convert to microfarads:
$$C \approx 0.597 \, \mu \mathrm{F}$$ The capacitance of the capacitor is:
$\boxed{0.597 \, \mu \mathrm{F}}$