“The
release of atomic energy has not created a new problem. It has merely made more
urgent the necessity of solving an existing one.”
– Albert
Einstein
Questions
(1) and (2) given below were asked in the National Eligibility cum Entrance Test
(NEET) conducted in July 2016. Question (3) is a modification of question (2).
Here are
the questions with their solution:
(1) A filament
bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works
perfectly and the bulb consumes 500 W. The value of R is :
(1) 26 Ω
(2) 13 Ω
(3) 230 Ω
(4) 46 Ω
The
resistance Rf of the
filament lamp related to its power P and
voltage V as
Rf
= V2/P = 1002/500 = 20 Ω
The
current through the filament lamp is P/V
= 500/100 = 5 A
Since the
main supply voltage is 230 volt and the total resistance in the series circuit
is (20 + R) ohm, we have
230 volt/(20 + R) ohm = 5 ampere
Therefore
(20 + R) = 46 from which R = 26 Ω
(2) The potential
difference (VA – VB) between the points A and B in the
given figure is
(1) + 6 V
(2) + 9 V 
(3) – 3 V
(4) + 3 V
The
voltage across 2 Ω resistor due to the current of 2A in it is 4 V. Similarly the voltage across 1 Ω
resistor is 2 V. These voltages and the battery voltage of 3 V are in conjunction and hence they add up. The potential
difference (VA – VB) between the points A and B is
therefore equal to (4V + 3 V + 2 V) = 9 V.
(3) If the battery voltage in question no.(2) is reversed as shown in the adjacent
figure, the potential difference (VA – VB)
between the points A and B will be
(1) + 12 V
(2) + 9 V 
(3) – 6 V
(4) + 3 V
In this case the voltages across the 2 Ω
resistor and the 1 Ω resistor are in conjunction while the battery voltage is in
opposition. Therefore, the potential difference (VA – VB)
between the points A and B is equal to (4V – 3 V + 2 V) = 3 V.
