SESSION:10

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1)       How do we decrease the slew rate of an op-amp?

Slew rate of an Operational Amplifier is the rate of change of output when there is a change in input signal. For s sinusoidal input the slew rate of Op-Amp is directly proportional to f and V. So either by reducing V or f SR can be decreased. Further you can use Capacitor for reducing slew rate.

2)       Which is more efficient to use 220v or 110v?

It deepens where to use. For lighting a Bulb we will use 220 V whereas foe connection of energy meter, we will take secondary voltage of Potential transformer which is 110 V. Also for a given power if you decrease the voltage current will increase and you then need to have more current carrying capacity element which costs more. 

3)       Why does voltmeter reading decrease, ammeter reading increase, when resistance is decreased?

Suppose the fixed resistance is R1 and voltage is V.

So current I = V/(R+R1)

Volatge measured by Voltmeter V1= IR=VR/(R+R1)

When the resistance R is decreased to value, say R2, current measured by Ammeter I'= V/(R2+R1)

So, Voltage measured by Voltmeter V2 =  I'R2 = VR2/(R2+R1)

So, V1/V2  = R(R2+R1)/R2(R+R1)

Lets assume R=20, R2=10 and R1=50

So, V1/V2 = 20*60/10*70 <1

which means V1<V2, hence voltmeter reading decreases.

Similarly, I'/I = (R+R1)/(R2+R1) = 70/60>1

so I'>I , hence current increases.

4)       What is current source?

A current source is a source of supply which produces constant current. It is basically an electronic circuit and it is assumed that infinite resistance is connected across the current source so that there is only one path for current to flow and can thus give constant value of current. The most basic example of current source is a Transistor in its active region.

5)       What is power factor and how does it effect in AC systems?

There are three elements of Electrical System, Resistance, Inductance and Capacitance. Suppose you connect AC supply to a combination of Capacitor,Inductor and Resistor, then because of their property the current flowing through them will not be in phase with the applied voltage rather it will either lag behind or lead the applied voltage depending upon whether Inductance dominates or Capacitance dominates. The cosine of this lagging or leading angle is called Power Factor. Power Factor is directly related to electrical power. In fact electrical power P=VICos(Φ ), so if the value of CosΦ is more i.e. nearer to one, the power will be more.

6)       There are 4 terminals in the motor. How can I find which are field winding and which are armature winding?

For DC Motor, the lead which is connected to the carbon brush is Armature winding and another set is of course Field winding.

7)       When we can amplify the voltage or power, why can't we produce a few volts at plants and amplify it to mega volts or watts?

Thing is that as per law of conservation of energy, Energy can neither be created nor be destroyed rather it can be converted from one form to another. So we can't amplify power, otherwise it will violate the law of conservation of energy. Also suppose, if it is possible to amplify power then Efficiency would be Output/Input>1, so more than 100%.

Thing is that during amplification we do Biasing by connecting proper voltage and then set up a level ,i.e, Active region for Transistor and then apply signal to the input of Transistor which then oscillates about the set Active Region. So current will become Signal current+set point. So signal is amplified but mind that we used some source of power to get into the active region, so power output is still equal to power input.

8)       Why can’t the induced voltage be greater than applied voltage?

Voltage is induced in a coil because of change of magnetic flux through it. Suppose AC Voltage V is applied across the terminals of a Coil, so full voltage V will not be applied across the terminals as there will be some voltage drop in the resistance of coil and connecting lead. Therefore the equation will be V=Ir+Induced Voltage(Back emf), therefore it is clear that Back emf can't be greater then V. If back emf is greater than V then the machine is Generator and Generator is then feeding and we are not applying Voltage on it. 

9)       If amperage were defined by the human body (E/R=I), what is the minimum voltage that will kill a person if the electricity enters through the finger?

If you touch an live wire of ungrounded system, you won't get shock as current will not flow through the body. For getting sensible shock the current flowing through the human body should be near around 10mA. If the current flowing through the body is more than 10mA but less than 20mA then it is fatal. But if it is more than 20mA, there can be death. So it is current because of which our heart and muscles get contract.  If the voltage is more than 50 Volt then you will get shock that is why control circuit are of 48 V or 24 V.

10)    Which is the killer? Voltage or current?

Current is the KILLER. If you touch an live wire of ungrounded system, you won't get shock as current will not flow through the body. For getting sensible shock the current flowing through the human body should be near around 10mA. If the current flowing through the body is more than 10mA but less than 20mA then it is fatal. But if it is more than 20mA, there can be death. So it is current because of which our heart and muscles get contract. 

11)    How does a capacitor or an inductor filter out the AC ripple in a rectifier to give a pure DC voltage?

A capacitor does this by absorbing the excess of voltage and supplying it whenever it is required. This accounts for several charging and discharging cycles occurring in the capacitor.

Thus, this leads to clearing off the ripples by making the output waveform quite smooth.

In fact the capacitor bypasses the AC component.