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This doubt arose when I was thinking of the following simple case.

I have an inductor coil of resistance $1\ \Omega$ and self inductance $0.5 \text{ H}$, this is connected to a variable voltage supply whose voltage increases at a rate of $2 \text{ V/s}$.

So, the increase in current per second due to the increase in voltage will be $2 \text{ A/s}$ (as $R=1\ \Omega$). And induced emf will be $-L\frac{dI}{dt} =-0.5 × 2 = -1 \text{ V}$.

Now, if I switch on the circuit then at $t = 1/4 \text{ s}$, my external voltage still would be $0.5 \text{ V}$.

Here is where my doubt arises: here my external voltage at this instant is $0.5 \text{ V}$ but induced is $1 \text{ V}$ (as induced emf depends on rate of change of current not current itself). So what would happen here, would the current flow in the opposite direction or I am missing some concept here?

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You can't assume the current will increase at a rate of $2\text{ A/s}$ to find the induced emf because the inductor voltage (or alternatively, the induced emf) will oppose the current flow, and prevent the current from increasing at the same rate as if there were no inductor.

You need to write Kirchhoff's voltage law (or equivalently, Faraday's law) along the loop, which will give you $$L \frac{dI}{dt} + RI = v_s $$ where $v_s$ is the supply voltage. You need to solve this differential equation. You will find that the current will always flow out of the positive terminal of the voltage source, i.e. not flow in the opposite direction.

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  • $\begingroup$ But, when we take purely inductive circuit there we see when voltage initialy increasing sinosiudaly to its max. , current still laggs by 90° i.e is in negative direction of voltage . Here ,the only way the current would be in negative direction is inductor voltage is greater than applied voltage $\endgroup$ May 4, 2022 at 5:49
  • $\begingroup$ @JaydeepKalal I wasn't claiming that current can't instantaneously oppose the voltage in any circuit. It just doesn't happen in your circuit. Either way, you need to solve Kirchhoff's voltage law/Faraday's law to figure out what happens. $\endgroup$
    – Puk
    May 4, 2022 at 6:24

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