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I was solving a question and got confused about steady state current in circuit (image), in solution it is given that steady state current across inductor will be zero as all the initial current of inductor dies out, but I wanted to know whether dependent sources always acts as a resistor (absorb power) or it can also deliver power so that steady state current can become some finite value also? If both conditions are possible then what factor determine steady state current in circuits(image) like this? enter image description here

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whether dependent sources always acts as a resistor (absorb power)

Consider the below simple example. The equations are

\$\frac{d i_{L1}}{dt} = \frac{1}{L1} V_{CCVS1} = \frac{1}{L1} \cdot 1 \cdot i_{L1}\$

This leads to an exponentially increasing current which goes to infinity (if initial current is not zero); i.e. no steady state.

schematic

simulate this circuit – Schematic created using CircuitLab

whether dependent sources always acts as a resistor (absorb power) or it can also deliver power

Oscillators designs use negative resistance elements along with LC(R) circuits to deliver power to compensate for the power dissipation in the LC(R) circuit. The negative resistance is often modelled as a dependent source (eg. BJT models). A example from Wikipedia showing the dependent source in a BJT model is shown below.

BJT Pi model diagram from Wikipedia

or it can also deliver power so that steady state current can become some finite value also?

Flip the + & - symbols of the \$6 \cdot i_1 (t)\$ dependent source in your circuit and try to find the solution. Like the example shown above, the source will now try to increase the current in the coil; i.e. supply power to increase energy stored in the coil.

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  • \$\begingroup\$ Thanks, great answer! I understand that all three cases(finite, zero or infinite value) at steady states are possible in circuit consists of inductor and dependent source, so shall I conclude that if circuit like this that contains only one inductor and if we calculate R( thevnin) across inductor and if it turns out to be +ve then steady state will be zero and if comes out - ve then some finite value at steady state and if dependent source depends on current of inductor then it's steady state will be infinite? Or there are no such generalisation and I have to apply laplace transform method? \$\endgroup\$
    – user215805
    Commented Jul 17, 2020 at 4:08
  • \$\begingroup\$ if dependent source depends on current of inductor then it's steady state will be infinite. I am not sure you can conclude so when resistors are present. The dissipation in the resistors may exceed the power supplied by the source. \$\endgroup\$
    – AJN
    Commented Jul 17, 2020 at 4:14
  • \$\begingroup\$ \$L_1 \frac{d i_{L1}}{dt} = V_{CCVS1} - i_{L1} R = 1 \cdot i_{L1} - R \cdot i_{L1} = (1 - R) \cdot i_{L1}\$. Consider the simple example with an additional resistor R in series. Now the convergence or divergence of the current depends on the value R of the resistor. \$\endgroup\$
    – AJN
    Commented Jul 17, 2020 at 4:20
  • \$\begingroup\$ In the example shown in the answer, note that the CCVS depends on its own current. It acts like a normal resistor (but it can take on positive or negative values). (A normal resistor drops a voltage proportional current flowing through itself). \$\endgroup\$
    – AJN
    Commented Jul 17, 2020 at 4:33

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