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Lets say each of the loops have the same resistance \$R\$ and the emf of the battery is \$V\$.

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Case1 :
There is no second loop. Then if I close the switch for \$t\$ seconds, then the energy supplied by the battery is \$\frac{V^2}{R}t\$ joules.

Case2 :
There is second loop and we notice current briefly in the second loop when the switch is closed. Suppose this current is constant \$i\$ and exists for \$t'\$ seconds. Where does the energy \$i^2Rt'\$ dissipated in this second loop come from ? I can guess it must be supplied from the battery itself. But my textbook doesn't say anything about this additional energy drawn from the battery. Is there any formula for the energy drawn from the battery when a second loop is present ?

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In case 1, when you closed the switch, there will be a time period where the current ramps up to the steady state value. This is due to the loop inductance. It stores energy in the magnetic field. When you open the switch there will be a spark across the contacts (case 1) that dissipates the energy stored.

In case 2, at the instant the switch is closed, the current flowing will be different to case 1. This is because transformer action induces a secondary voltage and current will flow in loop 2 via the ammeter. So now you have a short period of time where energy is delivered into loop 2. This adds an extra current demand on loop 1 when the switch closes and is in addition to the ramping-up current seen in case 1.

Is there any formula for the energy drawn from the battery when a second loop is present ?

In case 1 with the resistance in loop 1 at zero then, when the switch closes there is a rising current (di/dt) = V.L i.e. a rearrangement of Faraday's law. This is altered into an exponentially rising current that ultimately limits at V/R when the loop has resistance.

The formula for this is: -

enter image description here

This current creates a proportional magnetic field and some fraction of that field couples with loop 2 so, this generates a voltage in loop 2: -

\$V = N\dfrac{d\Phi}{dt}\$ where \$\Phi\$ is the coupled magnetic flux and N= 1 in your example.

That induced voltage drives current through the effective secondary inductance, the loop resistance and the ammeter. All in series.

The voltage and current (in loop 2) are taking energy from loop 1 in addition to the inductive energy held in the magnetic field for case 1.

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The power dissipated by the resistor in the first (left) circuit is less than \$V^2/R\$ because there is additional voltage drop \$V_{ind}=L\frac{dI}{dt}\$ accross the line when current is switched on because of its inductance (which is, however, not shown in the circuit diagram as inductance).

This difference in power is radiated as electromagnetic energy and may (or may not) be picked up by the second circuit.

Note that you cannot apply KVL because the preconditions of lumped circuit approximation are not satisfied: in this case \$\frac{\partial B}{\partial t} \ne 0\$ outside of circuit elements.

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The electrons in the first loop push on the electrons in the 2nd loop, and as those move they push back on first-loop electrons, upping losses in the battery powered loop.

There is a E&M thinker --- Jefimenko --- who states "the use of turtles to model the universe is wrong. Its electrons all the way down."

OK. OK. Jefimenko did not state E&M worked exactly like that, but his book "Causality, Electromagnetic Induction and Gravitation" is worth a read.

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