## Sunday, October 29, 2006

### M.C.Q. on A. C. Circuits

Problems in alternating current circuits are generally interesting and if you have a thorough idea about the vector method (phasor method), most problems become quite simple. To illustrate the method, let us consider the following MCQ:
In a series LCR circuit, the voltages across the inductor and capacitor are 18V and 12V respectively. If the output of the alternating voltage source connected to this series LCR circuit is 10V, what is the voltage drop across the resistor?
(a) 6V (b) 8V (c)10V (d) 12V (e) 4V
In the vector method, alternating currents and voltages are treated as vectors. The currents and voltages used here can be peak values or RMS values. Let us take RMS values since all currents and voltages mentioned in alternating current problems are RMS values (by convention) unless specified otherwise. You should note that the current and voltage are in phase in the case of a resistor. But the voltage across an inductor leads the current by π/2 where as the voltage across a capacitor lags behind the current by π/2. This means that the voltage drops (vectors) across an inductor and a capacitor are oppositely directed. The resultant voltage drop across the series LC combination in the above question is therefore 18V-12V= 6V and this voltage vector leads the voltage (vector) across the resistor by π/2. The supply voltage vector (10V) is the vector sum of the voltage drop across the resistor and the net voltage drop across the LC combination. [All these vectors are shown in the figure in which the current in the circuit is considered as a vector I in the positive X-direction. Note that this current vector is common for all the components since the circuit is a series circuit.. The voltage drop (vector VR) across the resistor also is taken along the positive X-direction since the current and voltage are in phase in the case of a resistor. The voltage across the inductor is indicated by the vector VL which leads the current vector I by π/2. It is therefore shown in the positive Y-direction. The voltage across the capacitor is indicated by the vector VC which lags behind the current vector I by π/2. It is therefore shown in the negative Y-direction.]
Therefore we have,
10 = √(VR2 + 62) where VR is the voltage drop across the resistor. From this we obtain VR= 8V.
In general, the supply voltage (V) is given by
V = √[VR2 +(VL-VC)2]
Here is another simple question:
An inductor is connected in series with a resistor. When a 15V, 50 Hz alternating voltage source is connected across this series combination, the voltage drop across the inductor is 5V. The voltage drop across the resistor will be
(a) 15V (b) 20V (c) 10V (d) 2√10 V (e) 10√2 V
The voltage drop vector VR in the case of the resistor is in phase with the current vector. But the voltage drop vector VL in the case of the inductor leads the current vector by π/2. Hence we have two vectors of magnitudes VR volts and 5 volts with an angle π/2 between them. Their vector sum gives the supply voltage vector of magnitude 15 volts. Hence we have,
15 = √ [VR2 + 52], from which VR = √200 = 10√2 volts.
What do you grasp from the above discussion? Certainly this: You cannot add voltage drops in AC circuits as scalar quantities. You have to perform vector addition taking into account the phase relation between the current and the voltage in the components. Here you have to forget your habit of scalar addition you perform in the case of direct current circuits!
The following MCQ appeared in the Karnataka CET (Medical) 2002 question paper:
When 100V D.C. is applied across a coil, a current of 1A flows through it. When 100V, 50Hz A.C. is applied to the same coil, only 0.5 A flows. The inductance of the coil is
(a) 5.5 mH (b) 0.55 mH (c) 55 mH (d) 0.55 H
Instead of giving you the resistance of the coil, you are given the direct voltage and the resulting direct current to enable you to calculate the resistance: R = V/I =100/1 =100 Ω. When the alternating voltage is applied, we have I = V/Z = V/√(R2 + L2 ω2 ) so that 0.5 = 100/√[1002 + L2 ×(100π)2]. Squaring, 0.25 = 104/ (104 + 104π2L2). Since π2 = 10 (nearly), this reduces to 0.25 = 1/(1 + 10L2), from which L = 0.55henry = 55 mH.
Now consider the following multiple choice question:
A series LCR circuit is connected to a 10V, 1 kHz A.C. supply. If L= 40mH, R= 50Ω and the current in the circuit is 0.2A, the voltage drop across the capacitor (in volts) is approximately
(a) 10V (b) 15V (c) 20V (d) 25V (e) 50V

The value of the capacitor is not given and you are asked to find the voltage drop across it. This is a problem involving resonance, indicated by the current of 0.2A (=10/50) which is limited by the resistor only. At resonance, the voltage drop across the capacitor is equal (in magnitude) to the voltage drop across the inductor, which is ILω = 0.2X0.04 × 2π×1000 = 16π = 50V, nearly. Therefore, the correct option is (e).
Let us now discuss the following question which appeared in the AIIMS 2003 question paper:
A capacitor of capacitance 2μF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, the voltage across the capacitor will be
(a) 0.16V (b) 0.32V (c) 79.5V (d) 159V

This is an example of how simple a question can be sometimes. The voltage drop across the capacitor = I/Cω = (2×10-3 )/(2 ×10-6×2π×1000) =0.16V.
The following MCQ also is a simple one, but some of you may be tempted to give the wrong answer.
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is shared equally between the electric and magnetic fields is
(a) Q (b) Q/2 (c) Q/√3 (d) Q/√2
Energy in the electric field means the energy of the capacitor and the energy in the magnetic field means the energy of the inductor. The total energy output of the oscillator is the maximum energy stored in the capacitor (or inductor), which is equal to Q2/2C. When the energy is shared equally between the electric and magnetic fields, the energy of the capacitor = ½ × Q2/2C. If ‘q’ is the charge on the capacitor in this condition, we have,
q2/2C = ½ × Q2/2C, so that q = Q/√2.
To conclude this post, let us discuss one more question:
In a series LCR circuit the voltage across each of the components L,C, and R is 80V. The supply voltage and the voltage across the LC combination are respectively
(a) 80V, 160V (b) 240V, 0V (c) 160V, 0V (d) 80V, 160V (e) 80V, 0V
Since the voltage across the capacitor and the inductor are equal in magnitude, the circuit is at resonance. Since the voltages (vectors!) across these components are in opposition, the net voltage across them (LC combination) is zero. The entire supply voltage will appear across the resistor. This is equal to 80V, as given in the question. So, the correct option is (e).
Very simple questions may tempt you to be careless while answering. Don’t be careless while answering this simple question which appeared in H.P.P.M.T. 2005 question paper:
An ideal transformer has Np turns in the primary and Ns turns in the secondary. If the voltage per turn is Vp in the primary and Vs in the secondary, Vs/Vp is equal to
(a) 1 (b) Ns/Np (c) Np/Ns (d) (Np/Ns)2
You should note that the magnetic flux linked per turn is the same for the primary and the secondary, since the coupling coefficient between them is almost unity. The rate of change of flux per turn also is the same, which means the induced voltage per turn is the same for the primary and the secondary. Therefore, the correct option is (a).
The following M.C.Q. appeared in Kerala Medical Entrance 2006 test paper:
A 16 μF capacitor is charged to 20 volt potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is
(a) 0.2 A (b) 40 mA (c) 2 A (d) 0.4 A (e) 0.8 A
‘Pure’ 40 mH coil means the resistance of the coil is zero. So, there is no energy loss due to Joule heating and the entire initial energy of the capacitor is handed over to the inductor when the current in the circuit is maximum. Therefore we have, ½CV2 = ½ L I2 with usual notations so that I = √[CV2/L] = √[(16×10-6×400)/ (40×10-3)] = 0.4 A

#### 2 comments:

1. this is one real unique blog!!!

2. Thank you Bharat. Your blog has many specialities which one cannot normally expect from an engineer. Keep your tempo undwindled.