“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

*R*_{f }of the filament lamp related to its power*P*and voltage*V*as*R*

_{f }=

*V*

^{2}/

*P*= 100

^{2}/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 (V

_{A}– V_{B}) between the points A and B in the given figure is
(1) + 6 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 (V_{A}– V_{B}) 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 (V_{A}– V_{B}) between the points A and B will be
(1) + 12 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 (V

_{A}– V_{B}) between the points A and B is equal to (4V – 3 V + 2 V) = 3 V.