(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) The earth wire is used on electrical appliances as a safety device to prevent electric shock when short circuit occurs in such appliances. The dangerous current is then passed into thq ground through the earth wire because the eartIthas a lower resistance than the body.
(b) Light from a bulb connected to a single cell dim and eventually goes off because of the defects of polarization and local action.
(c)
(i) \(V^2\)\(_{\text{rms}}\) + \(V_c - L_L)^2\)
\(V^2\)\( _{\text{rms}}\) = \(\sqrt{V^2 + (V_c - V_L)}^2\)
= \(\sqrt{30^2 + (70 - 30)^2}\)
= \(\sqrt{900 + 1600}\)
= 50v
(ii) (\(\beta\)) = X\(_L = 2\pi fL\)
= 2 \(\pi \times \frac{500}{\pi} \times 0.007\)
= 7 \(\Omega\)
But X\(_c = \frac{1}{2 \pi fC} = \frac{1}{2 \pi \times \frac{500}{\pi} \times 0.001}\)
= L\(\Omega\)
z = \(\sqrt{R^2 + (X_L - X_c)^2}\)
= \(\sqrt{8^2 + (7 - 1)^2} = \sqrt{64 + 36}\)
= \(\sqrt{100}\)
= 10\(\Omega\)
I = \(\frac{v}{z} = \frac{50}{10}\)
= 5.0 A
(iii) (\(\gamma\)) P = I\(^2R = 5^2 \times 8\) = 200W
\(V_o = V_\text{rms} \sqrt{2} = 50\sqrt{2} = 70.7V\)
But V\(_osin2 \pi ft\)
= 70.7\(sin 2 \times 2 \times \pi \times \frac{500}{\pi} \times t\)
= 70.7sin 1000t
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