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542 questions
A ring having a mean diameter of 21 cm and a cross – section of 10 cm2 is made of two semicircular section of cost iron and cost steel respectively with each joint having reluctance equal to air gap of 0.2 mm as shown in figure. Determine the ampere turns required to produce a flux of 0.8 mWb. The relative permeability of cost iron and cost steel are 166 and 800 respectively. Neglect fringing and leakage effects.
The ring consists of two parallel magnetic circuits of semiconductor cross sections having reluctances \(R_C\) and \(R_S\) corresponding to cost iron and cost steel respectively. The reluctance of the joint,
Prove that current is purely resistive circuit is in phase with applied A.C. voltage and current purely capacitive circuit leads applied voltage by 900 and draw their waveform.
A pure resistive circuit is shown in Fig. 12(a) contain a pure resistance R connected across an AC sinusoidal voltage source. The instantaneous value of an alternating source voltage is given by
Derive the expression of quality factor of series R-L-C circuit at resonance.
The quality factor is , in general, defined as the ratio of resonant frequency to bandwidth. It is given by
A two element series circuit consumes 700 V of power and has power factor of 0.707 leading when energized by a voltage source of waveform \(v=141\sin\left(314t+30^{\circ}\right)\). Find out the circuit element.
The power dissipated by the coil is \(P_C=I_{rms}V_{rms}\ cos\ \phi\) For this problem \(V_{rms}=\frac{V_{peak}}{\sqrt2}=\frac{141}{\sqrt2}=100\ V\)
The galvanometer shown in the circuit has a resistance of 5 Ω. Find the current through the galvanometer using Thevenin’s theorem.
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Fig.22(a)
Equivalent resistance of the circuit \(R_{th}=\frac{10\times15}{10+15}+\frac{12\times16}{12+16}=6+6.86=12.86\ \mathrm{\Omega}\)
Derive a mathematical expression for the average real power deliver by a single phase a.c. source with an e.m.f of \(e=\sqrt{2}E_m\sin\omega t\) when the source current is \(i=\sqrt{2}I_m\sin\left(\omega t-\theta\right)\). Define power factor of an a.c. circuit. State the major disadvantage of poor power factor.
\(e=\sqrt2E_m\sin{\omega t}\) … (1) \(i=\sqrt2I_m\sin{(\omega t-\theta)}\) …(2)
Derive an expression for the resonant frequency of a parallel circuit, one branch consisting of a coil of inductance L and resistance R and the other branch of capacitance C.
The series RL circuit with parallel C is given in Fig. 19. The designed circuit consists of resistance R, inductance L are in series with parallel configuration of capacitance C. Let us consider the current I flowing through the RLC circuit by the a.c. voltage source with constant value of V.
Explain
In this transformation any resistance in delta network can be found out if the resistances of corresponding star network are known. Here the branch resistances in the star connections are
A circuit receives 50 A current at a power factor of 0.8 lag from a 250, 50 Hz, 1-ph A.C. supply. Calculate the capacitance of the capacitor which is required to be connected across the circuit to make the power factor unity.
From the given data, we have \(cos\ \theta=0.8\), Or \(\theta=\cos^{-1}{0.8}={36.86}^0\)
Find the current in each branch of the network using Kirchhoff’s law
Fig.23
Applying KVL in mesh I, II and II respectively \(-2I_1-3I_2+{7I}_3=12\) … (1)
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