In an ideal step up transformer with a constant resistance attached to the secondary coil, how is energy conserved and ohms law followed at the same time?
$$\frac{V_p}{V_s}=\frac{N_p}{N_s}=\frac {I_s}{I_p}$$
For constant $V_p$ and $N_p$ ,
$$V_s \propto N_s$$
If total resistance to secondary coil is constant(including wires) , $V_s \propto I_s$ according to ohm's law.
But from first relation, $V_s \propto \frac {1} {I_s}$
What's happening here?
Best Answer
Actually, no. We have:
$N_s = N_p \dfrac{I_p}{I_s}$
So, $V_s \propto \frac {1} {I_s}$ is not true since $I_p$ is not constant.
The secondary voltage and current are proportional by Ohm's Law:
(1) $V_s = R_L I_s$
The power is conserved:
(2) $V_p I_p = V_s I_s = R_L I^2_s$
The secondary current is related to the primary current:
(3) $I_s = \dfrac{N_p}{N_s}I_p$
Thus:
$V_p I_p = R_L (N_p/N_s)^2 I^2_p$
$\rightarrow \dfrac{V_p}{I_p} = (N_p/N_s)^2R_L$