Q4GATE 2023MCQ

The four stator conductor $\left(A, A^{\prime}, B\right.$ and $\left.B^{\prime}\right)$ of a rotating machine are carrying DC currents of the same value, the directions of which are shown in figure (i). The rotor coils $a-a^{\prime}$ and $b-b^{\prime}$ are formed by connecting the back ends of conductor ' $a$ ' and ' $a^{\prime}$ ' and ' $b$ ' and ' $b^{\prime}$ ', respectively, as shown in figure (ii). The e.m.f. induced in coil $a-a^{\prime}$ and coil $b-b^{\prime}$ are denoted by $E_{a-a^{\prime}}$ and $E_{b-b^{\prime}}$ ; respectively. If the rotor is rotated at uniform angular speed $\omega \mathrm{rad} / \mathrm{s}$ in the clockwise direction then which of the following correctly describes the $\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}$ and $\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}$ ?

🚀 Solve in Practice Mode

📖 Explanation

Since, a-a' and b-b' are placed at GNA (Geometric Neutral Axis) therefore, $\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0$

Q8GATE 2019NAT

The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible. When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla(round off to three decimal places) calculated in the air gap is __________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} l_{ag} &=0.2 cm \\ l_m&= 40cm\\ B_0 &= 1 \text{Tesla}\\ \phi &= \frac{mmf}{\Re }\\ &\text{For the same mmf,} \\ \phi &\propto \frac{1}{\Re } \\ \text{In case-1:} &\\ \Re _1&=\frac{l_m}{\mu _0A} \\ &= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ \text{as} \;\;\mu _0=0& \\ \therefore \;\;\Re _1&= \frac{0.2 \times 10^{-2}}{\mu _0 A}\\ \text{In case-2:} &\\ \Re _2&= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ &=\frac{0.2 \times 10^{-2}}{\mu _0 A}+\frac{40 \times 10^{-2}}{1000\mu _0 A}\\ &=\frac{0.24\times 10^{-2}}{\mu _0 A} \end{aligned}

As flux $\phi \propto B$ for uniform cross section area.

\begin{aligned} \therefore \; B_2&=\frac{B_1 \times \Re _1}{\Re _2} \\ B_2&= \frac{1 \times 0.2 \times 10^{-2}/\mu _0 A}{0.24 \times 10^{-2}/\mu _0A}\\ &= 0.833T \end{aligned}

Q10GATE 2018MCQ

A transformer with toroidal core of permeability $\mu$ is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius $r\lt \lt R$ , and neglecting any leakage flux, the best estimate for the mean radius R is

🚀 Solve in Practice Mode

📖 Explanation

Since we know reluctance of core: $\Re =\frac{l_c}{\mu a}$ Here, $l_c=$ mean core length, a= area of core reluctance,

\begin{aligned} \Re &=\frac{2 \pi R}{\mu \times \pi r^2}\\ &=\frac{2R}{\mu r^2}\\ [\text{where} R &= \text{mean radius of core}] \\ \text{or}\;\; R&=\frac{\Re \mu r^2}{2} \\\text{and}\;\; \Re &= \frac{mmf}{flux}=\frac{N_pI_p}{\phi }\\ \text{Here,}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}e\;dt \\ [\text{where}\; e &\text{is primary generated voltage}] \\ \text{or}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}-(V_p)\;dt \\ \phi &= \frac{1}{N_p}\int_{0}^{t}V \cos \omega t \;dt\\ &= \frac{V}{\omega N_p}\sin \omega t\\ &\text{Now reluctance} \\ \Re &= \frac{N_p I \sin \omega t}{\frac{V}{\omega N_p} \sin \omega t}\\ &= \frac{N_p^2\omega I}{V}\\ & \text{So mean radius,}\\ R&=\frac{N_p^2 \times \omega I \times \mu r^2}{2V} \end{aligned}

Q11GATE 2017NAT

A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): $Z_{m}=(24.50+j12.75)\Omega$ (main winding), $Z_{n}=(24.50+j12.75)\Omega$ (auxiliary winding). Neglecting the magnetizing branch the value of the capacitance (in $\mu$ F ) to be added in series with the auxiliary winding to obtain maximum torque at starting is _______.

🚀 Solve in Practice Mode

📖 Explanation

Auxiliary winding impedance $=(24.5+j12.75)\Omega$ Let $X_c$ be the reactance of the capacitor connected in auxiliary winding

\begin{aligned} Z_a &= (24.5+j12.75-jX_c)\Omega \\ X_a &=(X_c-12.75),\; R_a=24.5\Omega \\ \therefore \;\; \phi _a &=\tan^{-1}\frac{X_c-12.75}{24.5}\;...(i) \\ Z_m&= 12.5+j15.75\\ &= 20.10\angle 51.56^{\circ}\Omega \\ \phi _m&= 51.56^{\circ} \end{aligned}

For maximum torque at starting, condition on $\phi _a$ and $\phi _m$ is given by $\left[ \phi _a=\frac{90^{\circ}-\phi _m}{2} \right]$ [latex] \phi _a=\frac{90-51.56}{2}=19.22^{\circ}};...(ii) [/latex] Comparing equation (i) and (ii),

\begin{aligned} \tan(19.22^{\circ}) &=\frac{X_c-12.75}{24.5} \\ 0.3486 &= \frac{X_c-12.75}{24.5}\\ X_c &=8.541+12.75 \\ \text{Capacitive reactance,}& \\ X_c &= 21.291\Omega \\ \frac{1}{2 \pi fc}&= 21.291\\ \Rightarrow \;C&= \frac{1}{2\pi \times 50 \times 21.291}\\ &= 149.51\mu F \end{aligned}

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