Question23. What is the capacitance of a capacitor that draws 150 mA when connected to a $100 \mathrm{~V}, 400 \mathrm{~Hz}$ voltage source?
Ans. 0,597 $\mu \mathrm{F}$ .

Studdy Solution

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 $C$ :
$C = \frac{I}{V \cdot 2\pi f}$

STEP 5

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:
$\boxed{0.597 \, \mu \mathrm{F}}$

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