Magnetically Coupled Circuits Network Topology And Filters Practice Questions

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.

Circuit in s-domain,

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

Loops =b-(N-1) 5=b-(8-1) b=12

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.

Number of KCL equations = n-1=5-1=4 Number of KVL equations =b-(n-1) = 7-(5-1)=3

$(R_{eq})_B=\frac{2.5}{a^2}$ $(2eq)_B=\left[ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right]$ (2eq)_A={\left[ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right]\}/b^2$$Z_{in}=\frac{2.5}{a^2b^2}+\frac{ j\omega 10^{-3}}{b^2}+\frac{1}{j\omega (10 × 10^{-5})}$$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$

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

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 $M=35\mu H$

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

Number of branches =b =6 No. of nodes = n= 4 No. of chords =b-(n-1) =6-(4-1)=3

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 $b_6$ $v_4+v_5-v_6=0$

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

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.

Number of links =B-(N-1) =B-N+1

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

Currently no explanation available