Tuesday, December 27, 2016

NEET 2016 Questions on Electric Circuits




“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.

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