## Saturday, March 02, 2013

### 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 (VR VS), in volts (VR and VS 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
VR = + 7 volt and
VS = + 3 volt
[Note that VR is the potential drop produced across the 7 Ω resistor connected between Q and R and VS is the potential drop produced across the 3 Ω resistor connected between Q and S].
Therefore (VR VS) = + 4 volt