Physics Paper 2, MAy/June. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main

Weakness/Remedies
Strength

Question 14

1.  Explain briefly the purpose of earthing an electrical appliance
2.  Why does the light from a bulb connected to a simple cell dim and eventually goes

off after a while?

1. A coil of inductance 0.007 H, a resistor of resistance 8Ω  and a capacitor of capacitance 0.001 F are connected in series to an a.c. source of frequency

Hz.  If the r.m.s voltages across the coil, the resistor and the capacitor are
30V,30V and 70V respectively.

1.  Draw a vector diagram to illustrate the voltage across the components in the circuit
2. Calculate the:

( )  r.m.s. voltage of the source;
(  r.m.s. current in the circuit;
power dissipated in the circuit.

1.  Write down the sinusoidal equation for the r.m.s. voltages, V,

in terms of the time, t.

_____________________________________________________________________________________________________
Observation

Part (a)  A lot of candidates attempted this question.  Many could not answer the question correctly.  Some wrote about lighting conductor instead of earthing.

Part (b)  This question was well attempted by majority of the responding candidates
Performance was above average.

Part (c)   This part was fairly attempted by the candidates.  Some candidates however lost marks due to their carelessness e.g
-   not putting arrows on the lines
-   length of the lines not relative to the magnitudes of the values they represent
-   right angles not indicated.
Many candidates could not give the correct sinusoidal equation for V in terms of time t.

The expected answers are:
(a)       The ground wire is a safety device
In the event of a short circuit occurring in the appliance the dangerous current,
instead of passing through our body  when we touch the appliances, gets sent
safely to the earth (ground)  through the grounding  wire because the ground has
much lower resistance than our body does .

(b)        Because they have two basic defects namely
-   polarization
-   local action

(c)(i)

Correct vector diagram
Correctly relative magnitude

(c) (i)  Frequency of the secondary output  =  50Hz
( V2rms    = VR2     + (Vc –VL) 2
=    302   + (70 – 30)2
=     (900 + 1600) =  2500
Vrms =  50V

(ii)     (   XL     =   2
= 7Ω
Xc  =

=              =     1 Ω

Z  =         R2    + (XL   - Xc)2

82    + (7 – 1)2                  =    64 + 36

=   10 Ω

I  =

=                 =    5.0 A

)         P =  I2R
=   52 x  8
=    200 W

(iii)    Vo   =   Vrms
=   50 x
=  70.7V

2  =  2 x
=   1000t
V  =  Vo  Sin  2

V = 70.7 sin 1000t