GATE EE · Electric Circuits

Magnetically Coupled Circuits Network Topology And Filters – GATE EE Electric CircuitsPractice Questions & PYQs

Generate GATE-level questions on Advanced Network topics. Focus on: 1. Mutual inductance, Dot convention, and Ideal transformers. 2. Network topology: Graph, Tree, Co-tree, Incidence, Tie-set, and Cut-set matrices. 3. Passive filters: Low-pass, High-pass, Band-pass, and Band-stop filters.

19 questions · 19 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024NAT

For the circuit shown in the figure, the source frequency is $5000 \mathrm{rad} / \mathrm{sec}$ . The mutual inductance between the magnetically coupled inductors is $5 \mathrm{mH}$ with their self inductances being $125 \mathrm{mH}$ and $1 \mathrm{mH}$ . The Thevenin's impedance. $Z_{\mathrm{th}}$ , between the terminals $P$ and $Q$ in $\Omega$ is _____ (rounded off to 2 decimal places).

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📖 Explanation

Given: $\omega=5000 \mathrm{rad} / \mathrm{sec}$

\begin{aligned} z_{\text {\in }} & =j X_{L 1}+\frac{(\omega m)^{2}}{j X_{L 2}+z_{L}} \\ j X_{L 1} & =125 \mathrm{mH} \times 5000=j 625 \Omega \end{aligned}

\begin{aligned} & \frac{-j}{\omega c}=\frac{-j}{5000 \times 50 \times 10^{-6}}=-j 4 \Omega \\ & j x_{L 2}=1 \mathrm{mH} \times 5000=+j 5 \Omega \\ & z_{\text {\in }}=j 625+\frac{\left(5000 \times 5 \times 10^{-3}\right)^{2}}{j 5-j 4}=j 625-j 625=0 \Omega \\ & z_{m}=\frac{4 \times 2}{4+2}+4=\frac{16}{3} \Omega=5.33 \Omega \end{aligned}

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Q2GATE 2022MSQ

The transfer function of a real system, $H(s)$ , is given as: $H(s)=\frac{As+B}{s^{2+}Cs+D}$ where A, B, C and D are positive constants. This system cannot operate as

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📖 Explanation

Put $s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}$ So, the system pass low frequency component. Put $s=\infty , H(\infty )=0$ For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.

Q3GATE 2021NAT

An air-core radio-frequency transformer as shown has a primary winding and a secondary winding. The mutual inductance M between the windings of the transformer is ____________ $\mu H$ .(Round off to 2 decimal places.)

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📖 Explanation

\begin{aligned} I_{1}&=\frac{5}{22}(p-p) \\ V_{0}&=j \omega M I_{1}=7.3=\left(2 \pi \times 10 \times 10^{3}\right) \times M \times \left(\frac{5}{22}\right) \\ M&=51.10 \mu \mathrm{H} \end{aligned}

Explanation Figure 2

Q4GATE 2021MCQ

The input impedance, $Z_{\in}\left ( s \right )$ for the network shown is

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📖 Explanation

Circuit in s-domain,

\begin{aligned} -S I_{1}+(4 s+5) I_{2} &=0 \\ \Rightarrow\qquad \qquad I_{2} &=\frac{s}{4 s+5} I_{1} \\ V_{1} &=(4+6 s) I_{1}-\frac{s^{2}}{4 s+5} I_{1} \\ \frac{V_{1}}{I_{1}} &=\frac{(4+6 s)(4 s+5)-s^{2}}{4 s+5} \\ &=\frac{24 s^{2}+30 s+16 s+20-s^{2}}{4 s+5} \\ Z_{\text {\in }} &=\frac{23 s^{2}+46 s+20}{4 s+5} \end{aligned}

Explanation Figure 2

Q5GATE 2019MCQ

The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120 $^{\circ}$ with respect to each other. The rms value of neutral current is

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📖 Explanation

$I_N=I_a+I_b+I_c$ $(I_N)_{rms}=100A$

Explanation Figure 1

Q6GATE 2018MCQ

The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is

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Q7GATE 2016NAT

The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is __________.

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📖 Explanation

The above circuit can be drown by transferring secondary circuit to primary side. $I=\frac{100V}{(8+10j-4j)\Omega }$ $\;\;=\frac{100V}{(8+6j)\Omega }$ So the rms value of I will be 10 A.

Explanation Figure 2

Q8GATE 2016MCQ

The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KCL equations and the number of independent KVL equations, respectively, are

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Q9GATE 2015MCQ

Find the transformer ratios a and b such that the impedance ( $Z_{\in}$ ) is resistive and equals 2.5 $\Omega$ when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s.

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📖 Explanation

$(R_{eq})_B=\frac{2.5}{a^2}$ $(2eq)_B=\left[ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right]$ $(2eq)_A={\left[ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right]\$ }/b^2Z_{in}=\frac{2.5}{a^2b^2}+\frac{ j\omega 10^{-3}}{b^2}+\frac{1}{j\omega (10 × 10^{-5})}Z_{in}=\frac{2.5}{a^2b^2}+j\left[ \frac{5 × 10^3 × 10^{-3}}{b^2} -\frac{1}{5 × 10^3 × 10 \times10^{-6}}\right]$ ;...(i) $From problem,$ Z_{\in}=2.5 ;;...(ii) $From equation (i) and (ii),$ a^2b^2=1\frac{5}{b^2}-\frac{1}{5 \times 10^{-2}}=0\Rightarrow ; b=0.5 \text{ and } a=2$

Explanation Figure 2

Q10GATE 2015NAT

Two identical coils each having inductance L are placed together on the same core. If an overall inductance of $\alpha$ L is obtained by interconnecting these two coils, the minimum value of $\alpha$ is ____.

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📖 Explanation

CASE-I: $L_{eff}=L_+L_2=2L,$ , a=2 CASE-II: $l_{eff}=\frac{L_1L_2}{L_1+L_2}=\frac{L^2}{2L}=\frac{L}{2}$ , a=0.5 CASE-III If both are diffferentially coupled then $L_{eff}=0$ Minimum value =0

Q11GATE 2014NAT

Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 $\mu$ H and 240 $\mu$ H. Their mutual inductance in $\mu$ H is _____.

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📖 Explanation

The two possible series connection are shown below: Let the mutual inductance be M (i) Additive connection, (ii) Substractive connection, $L_{eqv}=L_1+L_2-2M=240\mu H$ Thus, $L_1+L_2+2M=380 \mu H\;\;...(i)$ and $L_1+L_2-2M=240 \mu H\;\;...(ii)$ Solving equations (i) and (ii), we get: $4M=140\mu H$ $M=35\mu H$ Therefore, mutual inductance $M=35\mu H$

Explanation Figure 1

Explanation Figure 2

Q12GATE 2013MCQ

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage $V_{WX1}$ = 100 V is applied across WX to get an open circuit voltage $V_{YZ1}$ across YZ. Next, an ac voltage $V_{YZ2}$ = 100 V is applied across YZ to get an open circuit voltage $V_{WX2}$ across WX. Then, $V_{YZ1}/V_{WX1},V_{WX2}/V_{YZ2}$ are respectively,

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📖 Explanation

$V_{YZ_1}=100\times 1.25 \times 0.8 =100$ In second case when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding, Hence, $V_{WX_2}=\frac{100}{1.25}=80V$ $\Rightarrow \; \frac{V_{YZ_1}}{V_{WX_1}}=\frac{100}{100}$ $\frac{V_{WX_2}}{V_{YZ_2}}=\frac{80}{100}$

Q13GATE 2008MCQ

The number of chords in the graph of the given circuit will be

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Q14GATE 2007MCQ

The matrix A given below in the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes. Let $V=[v_{1} \; v_{2} ... v_{6}]^{T}$ denote the vector of branch voltages while $I=[i_{1} \; i_{2} ... i_{6}]^{T}$ that of branch currents. The vector $E=[e_{1} \; e_{2} \; e_{3} \; e_{4}]^{T}$ denotes the vector of node voltages relative to a common ground.

\begin{bmatrix} 1 & 1&1 & 0&0 & 0\\ 0& -1&0 & -1& 1& 0\\ -1 &0 & 0 & 0 &-1 & -1\\ 0&0 & -1&1 & 0 &1 \end{bmatrix}

Which of the following statement is true ?

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📖 Explanation

Convention: (when branch $b_j$ leaves away from node $N_i$ ) (when branch $b_j$ enters node $N_i$ ) Given node incidence matrix eg.: Branch $b_1$ enter $N_3$ and leaves $N_1$ The oriented network graph Using KVL, in loop containing branches $b_1,b_3$ and $b_6$ $v_1-v_6-v_3=0$ In loop containing branches $b_4,b_5$ and $b_6$ $v_4+v_5-v_6=0$

Explanation Figure 1

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Q15GATE 2002MCQ

A first order, low pass filter is given with R=50 $\Omega$ and C= $5\mu F$ . What is the frequency at which the gain of the voltage transfer function of the filter is 0.25?

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📖 Explanation

$T.F.=\frac{V_o}{V_i}=\frac{\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}$ $\;\;\;=\frac{1}{1+j\omega CR}$ $\therefore \; \; |Gain|=\frac{1}{\sqrt{1+(\omega CR)^2}}$ $\;\;0.25=\frac{1}{\sqrt{1+(\omega \times 5 \times 10^{-6} \times 50)^2}}$ On solving, $\omega =15.49 \times 10^3 r/s$ $\Rightarrow \;\; f=\frac{15.49}{2 \pi}=2.46 kHz$

Explanation Figure 1

Q16GATE 2002MCQ

In the circuit shown in figure, it is found that the input ac voltage $\left ( V_{i} \right )$ and current i are in phase. The coupling coefficient is $K=\frac{M}{\sqrt{L_{1}L_{2}}}$ , where M is the mutual inductance between the two coils. The value of K and the dot polarity of the coil P-Q are

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📖 Explanation

Input ac voltage and current will be in phase only at resonance condition, i.e. $X_C=X_L$ $|-j12|=|j8+j8+2k\sqrt{(j8)\times (j8)}|$ $12=8+8+16k$ $\Rightarrow \; k=-\frac{4}{16}=-0.25$ Hence, coupling will be opposite. Therefore, Dot will be at Q.

Q17GATE 2002MCQ

The graph of an electrical network has N nodes and B branches. The number of links, L, with respect to the choice of a tree, is given by

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Q18GATE 2001MCQ

Given two coupled inductors $L_{1}$ and $L_{2}$ , their mutual inductance M satisfies

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📖 Explanation

$M=K\sqrt{L_1L_2}$ where , K= coefficient of coupling $\because \; 0 \lt K \lt 1$ $\therefore \; M \leq \sqrt{L_1L_2}$

Q19GATE 2001MCQ

A connected network of $N\gt 2$ nodes has at most one branch directly connecting any pair of nodes. The graph of the network

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