Saturday, October 27, 2012

Multiple Choice Questions on Metre Bridge

“If I have seen a little further it is by standing on the shoulders of Giants.”
– Sir Isaac Newton

The metre bridge is basically a Wheatstone bridge, the condition of balance of which is conveniently used in measuring unknown resistances. Today we shall discuss a few questions (MCQ) involving metre bridge:
(1) Resistances R1 and R2 are connected respectively in the left gap and right gap of a metre bridge. If the balance point is located at 55 cm, the ratio R2/R1 is
(a) 4/5
(b) 5/4
(c) 9/11
(d) 11/9
(e) 6/7
The distance of the balance point from the left end of the wire in the metre bridge is 55 cm. Therefore the distance of the balance point from the right end of the wire is 45 cm. Thus we have
R1/R2 = 55/45
Or, R2/R1 = 45/55 = 9/11
(2) A 15 Ω resistance is connected in the left gap and an unknown resistance less than 15 Ω is connected in the right gap of a metre bridge. When the resistances are interchanged, the balance point is found to shift by 20 cm. The unknown resistance is
(a) 5 Ω
(b) 6 Ω
(c) 8 Ω
(d) 10 Ω
(e) 12 Ω

Initially the balance point will be at J1 as indicated in the figure. On interchanging the resistances the balance point will shift to J2 so that the length of the bridge wire between J1 and J2 is 20 cm. The balance points J1 and J2 must be equidistant from the mid point (50 cm mark) of the wire so that J1 is at 60 cm and J2 is at 40 cm.
Therefore we have
R/X =  60/40 = 6/4
Or, 15/X = 6/4
This gives X = 10 Ω
(3) Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a metre bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the shift in the balance point is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm
(e) 30 cm
Initially when the left gap and right gap contain 4 Ω and 6 Ω respectively, the condition of balance is
4/6 = L/(100 L) where L is the balancing length
This gives L = 40 cm.
When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the balancing length becomes 50 cm (since the gaps contain equal resistances).
Therefore the shift in the balance point is (50 cm – 40 cm) = 10 cm.

Tuesday, October 09, 2012

Questions Involving Charging of Capacitors

“Science distinguishes a Man of Honour from one of those Athletic Brutes whom undeservedly we call Heroes.”
– John Dryden

Today we will discuss a few multiple choice questions involving charging of capacitors. These questions will serve as practice questions for various entrance examinations including the National Eligibility Cum Entrance Test (NEET) for admission to MBBS and BDS Courses and the Joint Entrance Examination (JEE) for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions. Here are the questions:
(1) An initially uncharged capacitor of capacitance 100 μF is charged at a steady rate of 20 μC/s. What is the time (in seconds) required for the potential difference across the capacitor to build up to 12 V?
(a) 240 s
(b) 120 s
(c) 60 s
(d) 30 s
The final charge Q on the capacitor is given by
Q = CV = 100 μF×12 V = 1200 μC
Since the charging rate is 20 μC/s, the time t required for the potential difference across the capacitor to build up to 12 V is given by
t = 1200 μC/20 μC/s = 60 s

(2) A battery of emf V volt is connected to a combination of four identical capacitors (each of capacitance C farad) arranged as shown in the adjoining figure. What is the amount of energy  supplied by the battery to the capacitor combination?
(a) CV2/4
(b) CV2/2
(c) CV2/3
(d) 2CV2/3
The effective capacitance connected across the battery is C + (C/3) = 4C/3
The energy E of the capacitor combination on getting charged to V volt is given by
E = ½ × (4C/3) V2 = 2CV2/3
(3) A parallel plate capacitor with air as dielectric has capacitance C. It is charged fully using a battery of emf V. The battery is then disconnected and the separation between the plates of the capacitor is doubled. What is the final energy stored in the capacitor?
(a) ½ CV2
(b) CV2
(c) 3CV2/2
(d) 2 CV2
(e) CV2/4
The charge Q on the capacitor is unchanged since the battery is disconnected. The charge Q is given by
Q = CV ------------- (i)
When the separation between the plates is doubled, the capacitance is halved (C/2).
[Note that the capacitance of a parallel plate air capacitor is given by C = ε0A/d where A is the plate area and d is the separation between the plates.
The energy E stored in the capacitor is given by
E = Q2/2C1 where C1 = C/2
Substituting for Q from Eq. (i), we have
E = C2V2/(2C/2) = CV2, as given in option (b).
[Note that the increased energy of the capacitor is the result of the external work done for increasing the separation between the plates, overcoming the electrostatic attractive force between the plates]
(4) In the above question, suppose the battery remains connected to the capacitor while doubling the separation. What will be the final energy stored in the capacitor>
(a) ½ CV2
(b) CV2
(c) 3CV2/2
(d) 2 CV2
(e) CV2/4
In this case charges can flow through the wires connecting the battery to the capacitor. The capacitor will hold a smaller amount of charge (let us say, Q1) since the capacitance is decreased from C to C/2. We have\
Q1 = (C/2) V
The final energy (E1) stored in the capacitor is given by
E1 = Q12/2C1 = [(C/2)V]2/(2C/2) = CV2/4