Induced emf confusions

Suppose I have a simple ac circuit in which the ac source is connected to 3 resistors in series.After reading my 12th standard book I found that we can calculate potential difference across each resistor similar to as we do in a simple dc circuit. But here I am quite confused as I feel induced emf is much different from battery emf.Rectify me if I am wrong(and if I am wrong then please answer my this question-When the magetic field linked with a simple closed coil is changing an induced emf is set up but between which two point on the coil is this potential difference set up?)

You are correct that there are two kinds of EMF in the circuit: the EMF provided by the battery, and the "back EMF" in the other direction generated by the changing magnetic flux through the loop of wire that constitutes the circuit.

However, there are two big reasons why the back EMF is ignored here. The primary reason is that the back EMF is usually* tiny compared to the source EMF (i.e. the supply voltage). The back EMF $$\mathcal{E}_{ind}$$ from the loop of wire itself is found using the definition of inductance:

$$\mathcal{E}_{ind}=L\frac{dI}{dt}$$

In this case, we know that the AC current through the circuit looks like $$I=I_0\sin(\omega t)$$, for some low frequency $$\omega$$ (usually 50 or 60 Hz for grid power). Let's assume that the circuit is arranged along a circular loop of wire of radius $$a$$ (this is the configuration with maximum inductance per unit length, so it gives us a good "worst-case scenario"). Let's also assume that the wire itself has a circular cross-section with radius $$b$$ (this is fairly realistic). The inductance for such a wire loop can be estimated using the following formula**:

$$L=\mu_0 a\left(\ln\frac{8a}{b}-2\right)$$

If we insert the above two equations into the back EMF formula, we obtain:

$$\mathcal{E}_{ind}=\mu_0aI_0\omega\left(\ln\frac{8a}{b}-2\right)\cos(\omega t)$$

If our loop is 10 cm in radius and the wire itself is 2 mm thick (so it has a radius of 1 mm), then $$a=.1$$ m, $$b=.001$$ m, and by the above formula, $$L=5.9\times 10^{-7}$$ H. If we let $$\omega=2\pi\times 60$$ Hz and $$I_0=1$$ A, we get a maximum back EMF of around 0.2 mV, which is tiny compared to the usual AC supply voltage of 120-240 V.

As a secondary, more pedagogical reason, the amount of back EMF from the circuit itself depends on the geometry of the physical circuit. A longer loop of wire connecting the battery and resistors will generate a higher back EMF than a shorter loop of wire, and a more circular loop will generate a higher back EMF than a "flattened" loop. For the sake of simplicity, most textbooks use the "lumped-element model" to describe electric circuits, where wires connecting components are assumed to have zero resistance, inductance, and capacitance, making the physical circuit geometry irrelevant. In the lumped-element model, if you want to account for the properties of the wire itself, you insert them as an explicit resistor, inductor, and capacitor. Naturally, this makes the circuit much more complicated, so for educational purposes it is often ignored (since, as we have already shown, it's tiny and only makes things more complicated).

*The following discussion assumes that the AC circuit has a frequency that is low enough that the lumped-element circuit model is valid. At high frequencies, the impedance of most electronic components (including the wires themselves) drastically increases, and they all start radiating quite a bit of their energy away as electromagnetic radiation, so you can't really use the lumped-element model easily anymore.