BITSAT Questions (MCQ) on Direct Current Circuits

There are two ways to live your life. One is as though nothing is a miracle; the other is as though everything is a miracle.

– Albert Einstein

We shall discuss a few interesting multiple choice questions on direct current circuits, which appeared in BITSAT question papers.

The following question appeared in BITSAT 2009 question paper:

(1) In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B and C reads (100/3) V. Neglecting the internal resistance of the battery, the value of R is

(a) 100 KΩ

(b) 75 KΩ                                                                                  

(c) 50 KΩ

(d) 25 KΩ

Since the voltage drop axross B and C is (100/3) volt, which is 1/3 of the supply voltage, the effective resistance of the parallel combination of the voltmeter resistance R and the 50 KΩ resistor must be 25 KΩ.

[2/3 of the supply voltage is dropped across the 50 KΩ resistor in the gap AB and hence the effective resistance (in the gap BC) that drops 1/3 of the supply voltage must be 25 KΩ].

Since 50 KΩ  in parallel with 50 KΩ makes 25 KΩ, the internal resistance R of the voltmeter must be 50 KΩ [Option (c)].

[If you want to make things more clear, you may write the following mathematical steps:

The current I sent by the battery is given by

            I = 100/[50 + {(50×R)/(50+R)}]

In the above equation we have written the resistances in KΩ so that we will obtain the final answer in KΩ.

Since the voltage drop across B and C is (100/3) volt, we have

            100/3 = {(50×R)/(50+R)}× I

Or, 100/3 = {(50×R)/(50+R)}×100/[50 +{(50×R)/(50+R)}]

Rearranging, (50 + R) [50 +{(50×R)/(50+R)}] = 150 R

Or, 2500 + 50 R + 50 R = 150 R

This gives R = 50 and the answer is 50 KΩ since we have written resistances in KΩ].

The following question appeared in BITSAT 2005 question paper:

(2) Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 Ω is connected in series with the smaller of the two. The null point shifts to 40 cm.the value of the smaller resistance in ohms is:

(a) 3

(b) 6

(c) 9

(d)12

If P is the smaller resistance (Fig.) and Q is the larger resistance, we have

            P/Q = 20/80 = ¼ ………..   (i)

After connecting 15 Ω in series with P we have

            (P + 15)/Q = 40/60 = 2/3……….(ii)

On dividing Eq.(i) by Eq.(ii) we have

P/(P + 15) = 3/8

Therefore, 8P = 3P + 45 from which P = 9 Ω

The following question also appeared in BITSAT 2005 question paper:

(3) The current in a simple series circuit is 5 A. When an additional resistance of 2 Ω is inserted, the current drops to 4 A. The original resistance of the circuit in ohms was:

(a) 1.25

(b) 8

(c) 10

(d) 20

If the emf in the circuit is V volt and the original resistance of the circuit is R ohms we have

            V/R = 5 -----------------(i)

On inserting the additional resistance of 2 Ω we have

            V/ (R+2) = 4 -----------------(ii)

On dividing Eq.(i) by Eq.(ii) we have

            (R+2)/R = 5/4 

Or, 4R + 8 = 5R from which R = 8 Ω.

The following question also appeared in BITSAT 2008 question paper:

(4) A current of 2 A flows in an electric circuit as shown in the figure. The potential difference (V_R – V_S), in volts (V_R and V_S are potentials at R and S respectively) is

(a) – 4

(b) + 2

(c) + 4

(d) – 2

Since the two branches PRQ and PSQ contain equal resistances (10 Ω), the current gets divided equally at the junction P. The same current of 1 A flows through yhe branches. Taking Q as the reference point to measure the potentials at R and S we have

            V_R = + 7 volt and

            V_S = + 3 volt

[Note that V_R is the potential drop produced across the 7 Ω resistor connected between Q and R and V_S is the potential drop produced across the 3 Ω resistor connected between Q and S].

Therefore (V_R – V_S) = + 4 volt