(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) 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 after a while?
(c) A coil of incidence 0.007 H, a resistor of resistance 8 \(\Omega\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \(\frac{500}{\pi}\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;
(i) draw a vector diagram to illustrate the voltage across the components in the circuit.
(ii) Calculate the: (\(\alpha\)) r.m.s voltage of the source
(\(\beta\)) r.m.s current in the circuit;
(\(\gamma\)) power dissipated in the circuit.
iii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.
(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:
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:
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.
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:
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:
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.
