WAEC Physics 2013 Theory — Question 14

Model answer

(a) Earthing an electrical appliance is important because, in the event of a short circuit occurring in the appliance, the dangerous current, instead of passing through our body when we touch the appliance, gets sent safely to the ground through the earthing wire. This is because the ground has a much lower resistance than our body does. (b) The light from a bulb connected to a simple cell dims and eventually goes off after a while because the simple cell can only supply current for a short period of time. Two chemical processes that shorten the lifetime of the simple cell are: (i) Polarization — due to the chemical reactions taking place in the cell, hydrogen bubbles are formed at the copper plate. Some hydrogen bubbles rise to the surface and escape into thin air, while some remain at the copper plate. Accumulation of hydrogen bubbles around the copper plate acts as a barrier that increases the internal resistance of the cell. An increase in internal resistance results in a decreased output current, hence a decreased e.m.f. Polarization can be reduced by using a depolarizer. (ii) Local action — when the external circuit is removed, current ceases to flow and theoretically all chemical reactions within the cell stop; however, the zinc plate may contain impurities, forming small electrical cells within the zinc electrode, and current can flow between the zinc and its impurities. Thus chemical reactions continue even though the cell is not connected to a load; this is referred to as local action. (c) Given: Inductance L=0.007H, resistance R=8Ω, capacitance C=0.001F, frequency f=500/π Hz. VL=30V, VR=30V, VC=70V. (i) The vector diagram shows VR along the current axis, VL leading by 90° and VC lagging by 90° (opposite direction to VL); the resultant voltage V is the vector sum, drawn from (VL−VC) and VR. (ii)(α) r.m.s voltage of the source: V²rms = V²R + (VC−VL)² = 30² + (70−30)² = 900+1600 = 2500; Vrms = √2500 = 50V. (β) r.m.s current I(rms) = Vrms/Z, where Z=√(R²+(XL−XC)²). XL=2πfL=2×π×(500/π)×0.007=7Ω. XC=1/(2πfC)=1/(2×π×(500/π)×0.001)=1Ω. Z=√(8²+(7−1)²)=√(64+36)=√100=10Ω. I(rms)=50/10=5A. (γ) Power dissipated P = I²R = 5²×8 = 200W. (iii) Sinusoidal equation for r.m.s voltage: V = Vm sin(2πft), where Vm = Vrms×√2 = 50×√2 = 70.7V. So V = 70.7 sin(2π×(500/π)×t) = 70.7 sin(1000t).