(a) Explain briefly the purpose of earthing electrical appliance. (b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off a...

(a) Purpose of earthing an electrical appliance

The earth wire connects the metal casing of an appliance to the ground through a low-resistance path. If a fault causes the live wire to touch the casing, a large current flows through the earth wire rather than through a person touching the appliance. This causes the fuse to melt or the circuit breaker to operate, reducing the risk of electric shock.

(b) Why a bulb connected to a simple cell dims and eventually goes out

A simple cell suffers from polarisation and local action.

• During polarisation, hydrogen bubbles collect on the copper electrode. This produces a back e.m.f. and increases the effective internal resistance of the cell, so the current in the bulb decreases.
• Local action causes zinc to be used up by unwanted chemical reactions within the cell.

As the current falls, the bulb becomes dimmer. Eventually the cell can no longer provide enough current, so the bulb goes out.

(c) Series LCR circuit

The frequency is \(f=\dfrac{500}{\pi}\,\text{Hz}\), so the angular frequency is:

\[ \omega=2\pi f=2\pi\left(\frac{500}{\pi}\right)=1000\,\text{rad s}^{-1}. \]

Important check on the data: the stated component voltages are not consistent with the stated values of \(L\), \(C\), and \(R\). From the resistor voltage, \(I=20/8=2.5\,\text{A}\). This would predict \(V_L=I\omega L=17.5\,\text{V}\) and \(V_C=I/(\omega C)=2.5\,\text{V}\), rather than \(30\,\text{V}\) and \(70\,\text{V}\). Therefore the reference calculation giving \(50\,\text{V}\), \(5.0\,\text{A}\), and \(200\,\text{W}\) cannot follow from all the values in the question.

The calculations below use the directly stated r.m.s. voltages across the components, together with \(R=8\,\Omega\). This is the internally consistent approach for the voltage phasor diagram.

(i) Voltage vector diagram

Take the current, and therefore \(V_R\), as the horizontal reference direction. \(V_L\) is \(90^\circ\) ahead of the current, while \(V_C\) is \(90^\circ\) behind it. Since \(V_C\) is larger than \(V_L\), the circuit is overall capacitive.

(ii) Calculations

R.m.s. source voltage

The resistor voltage is perpendicular to the net reactive voltage:

\[ V_{\text{rms}}=\sqrt{V_R^2+(V_C-V_L)^2} \] \[ V_{\text{rms}}=\sqrt{20^2+(70-30)^2} =\sqrt{400+1600} =\sqrt{2000} =44.7\,\text{V}. \]

R.m.s. current

\[ I_{\text{rms}}=\frac{V_R}{R} =\frac{20}{8} =2.5\,\text{A}. \]

Power dissipated

For an ideal inductor and capacitor, only the resistor dissipates average power:

\[ P=I_{\text{rms}}^2R =(2.5)^2(8) =50\,\text{W}. \]

(iii) Sinusoidal equation for the source voltage

The peak voltage is:

\[ V_0=\sqrt{2}\,V_{\text{rms}} =\sqrt{2}(44.7) =63.2\,\text{V}. \]

If the source voltage is chosen to have zero phase at \(t=0\), its instantaneous voltage is:

\[ v=63.2\sin(1000t)\,\text{V}. \]

Examination reminder: In a series LCR circuit, do not add \(V_R\), \(V_L\), and \(V_C\) as ordinary numbers. \(V_L\) and \(V_C\) act in opposite vertical directions on the phasor diagram, so first find their difference, then combine it with \(V_R\) using Pythagoras.