Question 13 Marks
Obtain an expression for average power dissipated in a purely resistive A.C. circuit. Define power factor of the circuit and state its value for purely resistive A. C. circuit.
Questions
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Question 23 Marks
An emf of \(91 mV\) is induced in the windings of a coil, when the current in a nearby coil is increasing at the rate of \(1.3 A / s\), what is the mutual inductance \(( M )\) of the two coils in \(mH\) ?
Question 33 Marks
Assuming expression for impedance in a parallel resonant circuit, state the conditions for parallel resonance. Define resonant frequency and obtain an expression for it.
Question 43 Marks
Explain self induction and mutual induction.
Question 53 Marks
Prove theoretically, the relation between e.m.f. induced and rate of change of magnetic flux in a coil moving in a uniform magnetic field.
Answer
View full question & answer→Let us consider a rectangular wire loop PQRS of width l, and its plane perpendicular to a uniform magnetic field. The loop is being pulled out of magnetic field at a constant speed v. At any instant, let 'x' be the length of loop in the magnetic field. As the loop moves towards right, the area of the loop inside the field changes by dA=ldx=lvdt. So the change in magnetic flux is given by
\(d \phi= BdA = Blvdt\)
so \(\frac{ d \phi}{ dt }=\frac{ Blvdt }{ dt }= Bvl\)
Its direction is given by Fleming's left and rule.
So, the forces \(F_1\) and \(F_2\) on wires \(P R\) and \(Q S\) respectively are equal in magnitude and opposite in direction and have the same line of action. Hence they balance each other. There is no force on the wire RS as it lies outside the field. The force \(\overrightarrow{ F _3}\) on the wire \(PQ\) has magnitude \(F _3= IBL\) directed towards left. To move the loop with constant velocity \(\overrightarrow{ v }\) an external force must be applied. So work done by the external agent is given by
\(dw = Fdx =- IlBdx =- IBdA\)
\(=- Id \phi_{ m }\)
So rate of doing work \(P =\frac{ dw }{ dt }\)
\(= I \left(-\frac{ d \phi_{ m }}{ dt }\right)\)
but \(P = EI\)
so \(E =-\frac{ d \phi_{ m }}{ dt }\)
\(d \phi= BdA = Blvdt\)
so \(\frac{ d \phi}{ dt }=\frac{ Blvdt }{ dt }= Bvl\)
Its direction is given by Fleming's left and rule.
So, the forces \(F_1\) and \(F_2\) on wires \(P R\) and \(Q S\) respectively are equal in magnitude and opposite in direction and have the same line of action. Hence they balance each other. There is no force on the wire RS as it lies outside the field. The force \(\overrightarrow{ F _3}\) on the wire \(PQ\) has magnitude \(F _3= IBL\) directed towards left. To move the loop with constant velocity \(\overrightarrow{ v }\) an external force must be applied. So work done by the external agent is given by
\(dw = Fdx =- IlBdx =- IBdA\)
\(=- Id \phi_{ m }\)
So rate of doing work \(P =\frac{ dw }{ dt }\)
\(= I \left(-\frac{ d \phi_{ m }}{ dt }\right)\)
but \(P = EI\)
so \(E =-\frac{ d \phi_{ m }}{ dt }\)
Question 63 Marks
Obtain an expression for average power dissipated in series LCR A.C. circuit. Hence obtain an expression for power factor of the circuit.
Question 73 Marks
Explain the working of a transistor as a switch.
Answer
View full question & answer→Working of transistor as a switch:
a) A base-biased N-P-N common emitter transistor is as shown in figure (a).
b) On applying Kirchhoff ’s law in input circuit, we get,
$V_{BB} − I_BR_B = V_{BE}$
$\therefore V_{BB} = V_{BE} + I_BR_B$
c) Considering VBB as the d.c input voltage $‘Vi’$ then the above equation may be written as $V_i = V_{BE} + I_BR_B ….(i)$
Applying Kirchhoff’s law in output circuit, we get,
$V_{CE} = V_{CC} − I_CR_L$
d) Considering V_{CC} as the dc output voltage $‘Vo’$ then the above equation may be written as,
$V_o = V_{CC} − I_CR_C ….(ii)$
As Vi is increased from $0$ to $0.6 V$ (for silicon transistor),$ I_C = 0$ and the transistor is said to be in the cut-off state.
\therefore I_C = 0
From equation (ii),
V_o = V_{CC} = constant
f) As $‘V_i’$ is increased from $0.6 V$ to $1 V$, transistor will be in active state therefore IC increases linearly. From equation (ii) $‘V_o’$ decreases linearly. The transistor in this range is called in active state.
g) If $‘V_i’$ is increased further, $‘V_o’$ becomes non-linear and decreases further and tends to become zero, though its value nearly reaches zero. The transistor in this range is said to be in saturation state. The above characteristics are shown in figure (b).
h) When $V_i$ is low $(< 0.6 V), V_o$ is high and when $V_i$ is high $(> 1 V), V_o$ is low. By defining low voltage state as cut off state and high voltage level as saturation state, transistor can be used as a switch.
i) A low voltage input keeps the transistor in cut-off region (non-working) and the transistor is said to be switched off.
j) A high voltage input keeps the transistor in saturation state (working) and the transistor is said to be switched on.
a) A base-biased N-P-N common emitter transistor is as shown in figure (a).
b) On applying Kirchhoff ’s law in input circuit, we get,
$V_{BB} − I_BR_B = V_{BE}$
$\therefore V_{BB} = V_{BE} + I_BR_B$
c) Considering VBB as the d.c input voltage $‘Vi’$ then the above equation may be written as $V_i = V_{BE} + I_BR_B ….(i)$
Applying Kirchhoff’s law in output circuit, we get,
$V_{CE} = V_{CC} − I_CR_L$
d) Considering V_{CC} as the dc output voltage $‘Vo’$ then the above equation may be written as,
$V_o = V_{CC} − I_CR_C ….(ii)$
As Vi is increased from $0$ to $0.6 V$ (for silicon transistor),$ I_C = 0$ and the transistor is said to be in the cut-off state.
\therefore I_C = 0
From equation (ii),
V_o = V_{CC} = constant
f) As $‘V_i’$ is increased from $0.6 V$ to $1 V$, transistor will be in active state therefore IC increases linearly. From equation (ii) $‘V_o’$ decreases linearly. The transistor in this range is called in active state.
g) If $‘V_i’$ is increased further, $‘V_o’$ becomes non-linear and decreases further and tends to become zero, though its value nearly reaches zero. The transistor in this range is said to be in saturation state. The above characteristics are shown in figure (b).
h) When $V_i$ is low $(< 0.6 V), V_o$ is high and when $V_i$ is high $(> 1 V), V_o$ is low. By defining low voltage state as cut off state and high voltage level as saturation state, transistor can be used as a switch.
i) A low voltage input keeps the transistor in cut-off region (non-working) and the transistor is said to be switched off.
j) A high voltage input keeps the transistor in saturation state (working) and the transistor is said to be switched on.