(a)(i) What is meant by the root-mean-square value of an alternating current? (ii) Define impedance of an alternating current circuit. (b) An electrical dev...

(a)(i) Root-mean-square (r.m.s.) value of an alternating current

The r.m.s. value of an alternating current is the value of a steady direct current that would produce the same heating effect, at the same rate, in the same resistor.

For a sinusoidal current:

\[ I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}} \]

(a)(ii) Impedance

Impedance is the total opposition offered by an a.c. circuit to the flow of alternating current. It includes resistance and reactance due to capacitors and inductors.

\[ Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} \]

Its SI unit is the ohm, \(\Omega\).

(b) Capacitance required

The device must operate at its rated values, so the current in the series circuit is:

\[ I=\frac{P}{V}=\frac{60}{120}=0.50\ \text{A} \]

The voltage across the device is \(V_R=120\ \text{V}\), whereas the supply voltage is \(240\ \text{V}\). The capacitor voltage and the device voltage are \(90^\circ\) out of phase, so they must be added using Pythagoras, not by ordinary addition:

\[ V^2=V_R^2+V_C^2 \]

\[ V_C=\sqrt{240^2-120^2}=\sqrt{43200}=207.8\ \text{V} \]

The capacitive reactance is therefore:

\[ X_C=\frac{V_C}{I}=\frac{207.8}{0.50}=415.7\ \Omega \]

For a capacitor:

\[ X_C=\frac{1}{2\pi fC} \]

Therefore:

\[ C=\frac{1}{2\pi fX_C} \]

\[ C=\frac{1}{2(3.142)(50)(415.7)}=7.66\times10^{-6}\ \text{F} \]

\[ \boxed{C\approx7.7\ \mu\text{F}} \]

The expression \(C=\dfrac{X_C}{2\pi f}\) is incorrect: capacitance is inversely proportional to capacitive reactance.

(c)(i) Capacitance

The capacitance of a capacitor is the charge stored on either plate per unit potential difference across the capacitor:

\[ C=\frac{Q}{V} \]

Its SI unit is the farad (F).

(c)(ii) Capacitors in series

The diagram gives:

\[ C_1=\frac{5}{2}C_2 \]

For capacitors in series:

\[ \frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2} \]

\[ C_T=\frac{C_1C_2}{C_1+C_2} \]

Substitute \(C_1=\dfrac{5}{2}C_2\):

\[ C_T=\frac{\left(\frac{5}{2}C_2\right)C_2}{\frac{5}{2}C_2+C_2} \]

\[ C_T=\frac{\frac{5}{2}C_2^2}{\frac{7}{2}C_2}=\boxed{\frac{5}{7}C_2} \]

Capacitors in series carry the same charge. Since \(V=\dfrac{Q}{C}\), the smaller capacitance has the larger potential difference. Thus:

\[ \frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{C_2}{\frac{5}{2}C_2}=\frac{2}{5} \]

Also:

\[ V_1+V_2=2\ \text{V} \]

The voltage is divided in the ratio \(2:5\), giving:

\[ \boxed{V_1=\frac{4}{7}\ \text{V}} \]

\[ \boxed{V_2=\frac{10}{7}\ \text{V}} \]

Examination point: In a series capacitor circuit, charge is the same on each capacitor, but the potential difference is inversely proportional to capacitance.