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Here, all the initial conditions are : i1(0-)=0A and i2(0-)=0A When t≥0, the current i1(t) and i2(t) starts flowing. From my assumption, steady state value of i1(t) is 10A but what happens to i2(t)? Applying KVL, 10=i1(t)+di1(t)/dt - di2(t)/dt And i2(t)= di1(t)/dt-di2(t)/dt. The time constant τ of this Circuit is 2 sec. If at steady state di1(t)/dt =0 then di2(t)/dt=0;then i2(∞)=0A. According to my calculation, i1(t)=10-10e^(-0.5t) but what about i2(t)?How to figure out the i2(t) equation? Is it all zero? Please help me out with i2(t)? enter image description here I simulated the circuit and got these for 60Hz... enter image description here enter image description here enter image description here Both of them are same in simulation graph. How is it possible? Also this circuit can not be run at very high frequency. Please help me with i2(t).

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  • \$\begingroup\$ What values are the primary inductance and the mutual coupling in your simulation? The graphs suggest an "ideal transformer" not a real one. \$\endgroup\$ Feb 24 at 14:02
  • \$\begingroup\$ I took 1 H each case for simulation \$\endgroup\$ Feb 24 at 14:05
  • \$\begingroup\$ I don't have coupled inductor in my simulation software hence I tried using 1:1 ideal transformer \$\endgroup\$ Feb 24 at 14:15
  • \$\begingroup\$ Why have you used a 60Hz AC signal when the question clearly shows a DC supply? \$\endgroup\$
    – Finbarr
    Feb 24 at 14:43
  • \$\begingroup\$ Please help me with i1(t) and i2(t). If source is dc then what will be i2(t)? Because in steady state, i1 should be 10A. \$\endgroup\$ Feb 24 at 16:15
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I decided on something similar but different. Can you see why R/Lin= 2ms ? when Lp=1 , R=1

enter image description here

enter image description here

Your AC simulation lacks some spec so must be different from your expectations. What is Zin(f)?

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