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 R_f of the filament lamp related to its power P and voltage V as

            R_f = 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 (V_A – V_B) 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 (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

(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 (V_A – V_B) between the points A and B is equal to (4V – 3 V + 2 V) = 3 V.