GATE 2018 EE Q.26: solution and answer

GATE 2018 · EEQ.26 · 2 marks · MCQ

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

GATE 2018 EE question 26: A toroidal core with mean radius R (dashed mean-path circle, R drawn as an arrow from the centre to the dashed circle) and… — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
$\frac{μ V r ^{2} N _{P}^{2} ω}{I}$
(B)
$\frac{μ I r ^{2} N _{P} N _{S} ω}{V}$
(C)
$\frac{μ V r ^{2} N _{P}^{2} ω}{2 I}$
(D)
$\frac{μ I r ^{2} N _{P}^{2} ω}{2 V}$

Answer and solution

ANSWER

The mean radius is $R = \frac{μ I r ^{2} N _{P}^{2} ω}{2 V}$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2018 EE (IIT Guwahati)

Since $i_{S} = 0$ , the secondary carries no current, so the core is magnetised by the primary ampere-turns $N_{P} i_{P}$ alone.
Apply Ampere's circuital law along the mean circular path of radius $R$ : $H \cdot 2 π R = N_{P} i_{P} \Rightarrow B = μ H = \frac{μ N _{P} i _{P}}{2 π R}$ . Because $r ≪ R$ , this $B$ is taken as uniform over the cross-section.
The core flux through the circular cross-section of area $π r^{2}$ is $Φ = B π r^{2} = \frac{μ N _{P} r ^{2}}{2 R} I sin ω t$
Apply Faraday's law to the primary, neglecting leakage and winding resistance: $v_{P} = N_{P} \frac{d Φ}{d t} = \frac{μ N _{P}^{2} r ^{2} ω I}{2 R} cos ω t$ This has the same $cos ω t$ form as $v_{P} = V cos ω t$ . That fits a purely inductive magnetising current, which lags the voltage by $9 0^{\circ}$ . Equivalently, the primary self-inductance is $L = \frac{μ N _{P}^{2} r ^{2}}{2 R}$ and $v_{P} = L \frac{d i _{P}}{d t}$ .
Equate the amplitudes: $V = \frac{μ N _{P}^{2} r ^{2} ω I}{2 R}$ .
Solve for the mean radius: $R = \frac{μ I r ^{2} N _{P}^{2} ω}{2 V}$ This is option (D).

KEY IDEA

With the secondary open, the primary behaves as a toroidal inductor with $L = \frac{μ N _{P}^{2} ( π r ^{2} )}{2 π R} = \frac{μ N _{P}^{2} r ^{2}}{2 R}$ . Its voltage and current amplitudes satisfy $V = ω L I$ . Solving this for $R$ gives the answer.

WATCH OUT FOR

There are three easy slips. (1) Using the mutual inductance, which gives $N_{P} N_{S}$ as in option (B), when it is the PRIMARY voltage and current that are given. The self-inductance goes as $N_{P}^{2}$ . (2) Inverting the relation to $R \propto V / I$ as in options (A) and (C). A larger radius means a longer magnetic path and so a SMALLER inductance, so $R$ must grow with $I / V$ . (3) Dropping the factor 2 that comes from $π r^{2} / (2 π R)$ .

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 12 Magnetostatics questions, with notes

GATE 2018 EE Q.26: solution and answer

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