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
