RI 2004 Prelims P2 Q5
The figure below shows two lamps X and Y which are operating at normal brightness from a 20 V battery.
(a) Calculate
(i) the current from the battery.
(ii) the value of the resistance R.
(b) A copper wire is now connected between junctions C and D.
State and explain the effect on each of the lamps.
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Answer:
(a)
(i) Using P = IV, I = P/V
I = Ix + Iy (Sum of currents)
= (24 / 12) + (12 / 12)
= 2A + 1A
= 3A
(ii) V = IR
20 - 12 = 3 * R
R = 8/3 = 2.7 Ω (2 s.f.)
(b) Both lamps will not light up. This is because copper wire has an extremely low resistance when compared to the lamp's. As a result, the two lamps are short circuited, such that all the current (or charges) flow through the copper wire instead of flowing through the lamps.
