Question

An ideal transformer has ${\mathit{N}}_{\mathbf{1}}$ windings in the primary and ${\mathit{N}}_{\mathbf{2}}$windings in its secondary. If you double only the number of secondary windings, by what factor does (a) the voltage amplitude in the secondary change, and (b) the effective resistance of the secondary circuit change?

Short answer

(a)The voltage amplitude in the secondary coil is doubled.

(b) The resistance increases four times.

Step-by-step solution

Step-1: Formulas used  

Terminal voltage across the primary and the secondary coils are related to the number of turns by the equation

$\frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{V}}_{\mathbf{1}}}$

Also for a transformer,

${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}{\left({\displaystyle \frac{N_2}{N1}}\right)}^{\mathbf{2}}}{\mathbf{R}}$

Step-2: Calculations for secondary voltage

As shown by the equation in step 1, the voltage amplitude is directly proportional to the number of turns. So, when the number of turns is doubled, the voltage amplitude in the secondary coil is also doubled.

Step-3: Calculations for resistance

From the given equation in step-1

$\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}={\left(\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}\right)}^2$.

So, on doubling the number of turns, the resistance increases four times.

Hence, (a)The voltage amplitude in the secondary coil is doubled.

(b) The resistance increases four times.