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, 4

*R*+ 8 = 5*R*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
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