# In a transformer why do we consider the lost voltage to resistance negligible?

In my text book it says:
"The induced emf in the primary coil opposes the pd applied to the primary coil,$V_p$. Assuming the resistance of the primary coil is negligible, so all the applied pd acts against the induced emf in the primary coil, the applied pd is $V_p = \frac{\Delta \phi}{\Delta t}$."
Now I have some worries about this, consider a circuit containing only the primary coil and the ac supply. In this circuit the only resistance is across the wires in the coil, if this resistance is tiny we would expect that the current is huge since the product of the resistance and this current must be equal to the voltage. Now lets say we turn it off and add in the other components of our transformer and then switch it back on. Now we get a back emf and we get losses in voltage thanks to the resistance in our coil. Since all voltage must be lost across this coil we can say that:
$V_p -I_pR_p= \frac{\Delta \phi}{\Delta t}$
And we know $R_p$ is small but I dont see how we know $I_p$ is not a large number, because in order to know that we would have to determine somehow that $\frac{\Delta \phi}{\Delta t}$ is really close to $V_p$ or we'd need some kind of extra information. Because I would think it might be possible from just this information that there would be large losses to just like in the circuit with just the primary coil alone in it. I'm probably missing something and I'll post the page in the book so you can call me out on being stupid when it is inevitably right there. Thankyou! -Michael PAGE IMAGE HERE

• As you point out, that analysis relies on the assumption of negligible resistance in the winding. To someone who actually works with transformers, this assumption is so bad as to render the exercise useless, IMHO. Commented Apr 8, 2018 at 17:59

• Suppose nothing is connected to the secondary, so this might as well not be there. The relationship you seem not to be using is that $\frac{d \Phi}{dt}=-L \frac{dI}{dt}$, in which $I$ is the primary current. L is the $self inductance$ of the primary. $L$ depends mainly on the number of turns on the primary, and the cross-sectional area and material of the core. It is $very roughly$ a constant. Now you have an equation from which $I$ can be calculated: $$\mathscr{E}_{applied} -L \frac{dI}{dt}=IR$$. If $I= 0$ when $t=0$, this gives $$I=\frac{\mathscr{E}_{applied}}{R} (1-e^{\frac{Rt}{L}})$$ Commented Apr 9, 2018 at 11:55
• Please forget the end of my previous comment, the bit that starts "If $I=0$ when $t-0$". It would apply if we were applying a steady voltage to the primary, and we won't be doing that. I was daydreaming. Commented Apr 9, 2018 at 12:46
• Thank you, I might have to do some research as I haven't heard of the relationship you mentioned and I haven't seen L used anywhere before. But assuming L and R to be constants and treating the emf as alternating current ($\varepsilon =\varepsilon _0 sin(\omega t)$) we can solve the differential equation to get $I = \frac{\varepsilon_0 L}{R^2+\omega ^2 R^2} (Rsin(\omega t)+\omega L cos(\omega t))$. Now according to this equation an smaller R gives a way larger current. If we multiply the whole expression by R we get IR (lost voltage) and it looks like it increases as R decreases. Commented Apr 9, 2018 at 16:32