Question 23.6: A battery charger meant for a series connection of ten nicke...

Strategy and Solution for (a)
You would expect the secondary to have a small number of loops. Solving \frac{V_{ s }}{V_{ p }}=\frac{N_{ s }}{N_{ p }} for N_{ s } and entering known values gives

N_{ s }=N_{ p } \frac{V_{ s }}{V_{ p }}                    (23.32)

=(200) \frac{15.0 V }{120 V }=25.

Strategy and Solution for (b)
The current input can be obtained by solving \frac{I_{ s }}{I_{ p }}=\frac{N_{ p }}{N_{ s }} for I_{ p } and entering known values. This gives

I_{ p }=I_{ s } \frac{N_{ p }}{N_{ s }}                    (23.33)

=(16.0 A ) \frac{25}{200}=2.00 A.

Discussion
The number of loops in the secondary is small, as expected for a step down transformer. We also see that a small input current produces a larger output current in a step-down transformer. When transformers are used to operate large magnets, they sometimes have a small number of very heavy loops in the secondary. This allows the secondary to have low internal resistance and produce large currents. Note again that this solution is based on the assumption of 100% efficiency—or power out equals power in \left(P_{p}=P_{s}\right) —reasonable for good transformers. In this case the primary and secondary power is 240 W. (Verify this for yourself as a consistency check.) Note that the Ni-Cd batteries need to be charged from a DC power source (as would a 12 V battery). So the AC output of the secondary coil needs to be converted into DC. This is done using something called a rectifier, which uses devices called diodes that allow only a one-way flow of current.