**Capacitor Questions**

These questions are related to Capacitor Circuit, Capacitor Connections, Capacitive Reactance, and RC Circuit Time Constant which are are covered in detail here:

Capacitor in Series | Capacitors in Parallel

**1. Define capacitance.**

The ability of an electric circuit or component to store electric energy by means by means of an electrostatic field.

**2. Compare between an inductor and a capacitor the manner in which energy is stored.**

The capacitor stores energy in an electrostatic field, the inductor stores energy in a magnetic field.

**3. Common practical applications for capacitors list four.**

**1.** Power factor correction of an electrical system.

**2**. Improving torque in motors.

**3**. Filters in AC circuits.

**4.** Timing of control circuits

**4. The base unit used to measure capacitance is what?**

Farad (F)

**5. What three factors determine the amount of capacitance in a capacitor?**

**1.** Area of the plates

**2.** Type of dielectric

**3.** Spacing between plates

**6. What factors determine the voltage rating of a capacitor?**

The voltage rating of a capacitor indicates the maximum voltage that can be safely applied to its plates and depends on the insulation strength of its dielectric.

**7. To provide a total capacitance of 100µF, how would you connect two 50µF capacitors?**

In parallel.

**8. Calculate to total capacitance at a maximum voltage for two 220µF, 300-V capacitors connected in series.**

110µF.

**9. With a 25KΩ resistor connected in series with a 1,000µF capacitor and operated from a 12-VDC source:**

**a. **Calculate the RC time constant.

Let’s calculate it using the following formula:

$\tau =RC=25*{{10}^{3}}*1000*{{10}^{-6}}=\text{25seconds}$

**b. **When a voltage is applied to the circuit approximately how long will it take for the voltage across the capacitor to reach 12V?

125 seconds. Approximately, it will take 5-time constants which are equal to 125 seconds in this case.

**c. **If the fully charged capacitor is discharged through a 25Ω resistor, what would the value of the voltage across the capacitor be after the first 25 seconds of discharge?

4.416 V. After first time constant, the capacitor voltage decreases to 36.7 % of the fully charged voltage value.

**10. Define capacitance reactance.**

The opposition to AC current flow by a capacitor.

**11. When measuring capacitance reactance what is the base unit used?**

Ohm (Ω)

**12. Capacitance reactance is affected in what way by capacitance?**

As capacitance goes up capacitance reactance goes down

\[{{X}_{C}}=\frac{1}{2\pi fC}\]

**13. Capacitance reactance is affected in what way by frequency?**

As frequency goes up capacitance reactance goes down.

\[{{X}_{C}}=\frac{1}{2\pi fC}\]

**14. With a frequency of 60Hz and an AC voltage of 240 Volts applied to a 50µF capacitor. Calculate the amount of AC current flow in the circuit.**

First, let’s calculate reactance by the following formula:

\[{{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*50*{{10}^{-6}}}=53.1\Omega \]

Now, calculate current from the following formula:

\[I=\frac{V}{{{X}_{C}}}=\frac{240}{53.1}=4.52A\]

**15. A 10µF(C1) and a 40µF(C2) capacitor are connected in series to a 230V, 60Hz source. Calculate the value of the voltage drop across each capacitor.**

First, we will calculate total capacitance:

\[{{C}_{T}}=\frac{{{C}_{1}}*{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{10*40}{10+40}=8\mu F\]

Let’s use the following formulas to compute voltage drop across each capacitor:

\[\begin{align} & {{V}_{{{C}_{1}}}}=\frac{{{C}_{T}}}{{{C}_{1}}}*{{V}_{s}}=\frac{8}{10}*230=184V \\ & {{V}_{{{C}_{2}}}}=\frac{{{C}_{T}}}{{{C}_{2}}}*{{V}_{s}}=\frac{8}{40}*230=146V \\ & {{V}_{s}}={{V}_{{{C}_{1}}}}+{{V}_{{{C}_{2}}}}=184+46=230V \\\end{align}\]

**16. A 10µF(C1) and a 40µF(C2) capacitor are connected in parallel to a 280V, 60Hz source. **

**a. What is the capacitive reactance and current flow through C1?**

We will use the following formulas to compute reactance and current flow:

$\begin{align} & {{X}_{{{C}_{1}}}}=\frac{1}{2\pi f{{C}_{1}}}=\frac{1}{2\pi *60*10*{{10}^{-6}}}=265.258\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{{{C}_{1}}}}}=\frac{280}{265.258}=1.055A \\\end{align}$

**b. What is the capacitive reactance and current flow through C2?**

We will use the following formulas to compute reactance and current flow:

$\begin{align} & {{X}_{{{C}_{2}}}}=\frac{1}{2\pi f{{C}_{2}}}=\frac{1}{2\pi *60*40*{{10}^{-6}}}=66.314\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{{{C}_{2}}}}}=\frac{280}{66.314}=4.22A \\\end{align}$

**17. What type of power is associated with a capacitor called and in what units is it measured?**

Capacitive reactive power, which is measured in VARs