(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
Earthing means connecting the metal casing of an appliance to the ground through a low-resistance earth wire. Its purpose is safety: if a live wire becomes loose and touches the metal casing (a fault), the casing would otherwise become live and give a fatal shock to anyone who touches it. The earth wire provides a path of very low resistance straight to the ground, so the large fault current flows to earth (rather than through the user's body) and blows the fuse, cutting off the supply. Thus earthing protects the user from electric shock.
(b) Why the bulb of a simple cell dims and goes off
The light produced by a bulb connected to a simple (voltaic) cell gradually dims and finally goes off because of two defects of the simple cell:
Polarisation: hydrogen gas liberated during the chemical reaction collects as a layer of bubbles on the copper (positive) plate. This layer increases the internal resistance and also sets up a back e.m.f. that opposes the current, so the current falls.
Local action: impurities in the zinc plate form tiny local cells with the zinc, so the zinc is eaten away (and current is wasted) even when the cell is not in use, weakening the cell.
Together these defects steadily reduce the current, so the bulb dims and eventually goes out.
(c) Series R-L-C circuit on an a.c. source
Given: coil \(L = 0.007\,\text{H}\), resistor \(R = 8\,\Omega\), capacitor \(C = 0.001\,\text{F}\), frequency \(f = \dfrac{500}{\pi}\,\text{Hz}\). The r.m.s voltages are: across the coil \(V_L = 30\,\text{V}\), across the resistor \(V_R = 20\,\text{V}\), across the capacitor \(V_C = 70\,\text{V}\).
(i) Vector (phasor) diagram of the voltages
Taking the current \(I\) as the reference (horizontal) direction: \(V_R\) is in phase with the current, \(V_L\) leads the current by \(90^\circ\) (drawn upward) and \(V_C\) lags the current by \(90^\circ\) (drawn downward). The source voltage \(V\) is the resultant.
Phasor diagram: current is the reference direction; V_R is in phase, V_L leads by 90 degrees (up), V_C lags by 90 degrees (down). The resultant of the sides V_R = 20 V and (V_L - V_C) = -40 V gives the source voltage V = 44.7 V at a phase angle of 63.4 degrees below the current.
(ii) Calculations
(\(\alpha\)) r.m.s voltage of the source
From the phasor diagram, the resultant of the perpendicular components gives
Earthing means connecting the metal casing of an appliance to the ground through a low-resistance earth wire. Its purpose is safety: if a live wire becomes loose and touches the metal casing (a fault), the casing would otherwise become live and give a fatal shock to anyone who touches it. The earth wire provides a path of very low resistance straight to the ground, so the large fault current flows to earth (rather than through the user's body) and blows the fuse, cutting off the supply. Thus earthing protects the user from electric shock.
(b) Why the bulb of a simple cell dims and goes off
The light produced by a bulb connected to a simple (voltaic) cell gradually dims and finally goes off because of two defects of the simple cell:
Polarisation: hydrogen gas liberated during the chemical reaction collects as a layer of bubbles on the copper (positive) plate. This layer increases the internal resistance and also sets up a back e.m.f. that opposes the current, so the current falls.
Local action: impurities in the zinc plate form tiny local cells with the zinc, so the zinc is eaten away (and current is wasted) even when the cell is not in use, weakening the cell.
Together these defects steadily reduce the current, so the bulb dims and eventually goes out.
(c) Series R-L-C circuit on an a.c. source
Given: coil \(L = 0.007\,\text{H}\), resistor \(R = 8\,\Omega\), capacitor \(C = 0.001\,\text{F}\), frequency \(f = \dfrac{500}{\pi}\,\text{Hz}\). The r.m.s voltages are: across the coil \(V_L = 30\,\text{V}\), across the resistor \(V_R = 20\,\text{V}\), across the capacitor \(V_C = 70\,\text{V}\).
(i) Vector (phasor) diagram of the voltages
Taking the current \(I\) as the reference (horizontal) direction: \(V_R\) is in phase with the current, \(V_L\) leads the current by \(90^\circ\) (drawn upward) and \(V_C\) lags the current by \(90^\circ\) (drawn downward). The source voltage \(V\) is the resultant.
Phasor diagram: current is the reference direction; V_R is in phase, V_L leads by 90 degrees (up), V_C lags by 90 degrees (down). The resultant of the sides V_R = 20 V and (V_L - V_C) = -40 V gives the source voltage V = 44.7 V at a phase angle of 63.4 degrees below the current.
(ii) Calculations
(\(\alpha\)) r.m.s voltage of the source
From the phasor diagram, the resultant of the perpendicular components gives
