Respuesta :
Explanation:
Let [tex]N_p\ and\ N_s[/tex] are the number of turns in primary and secondary coil of the transformer such that,
[tex]\dfrac{N_p}{N_s}=\dfrac{1}{3}[/tex]
A resistor R connected to the secondary dissipates a power [tex]P_s=100\ W[/tex]
For a transformer, [tex]\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}[/tex]
[tex]V_s=(\dfrac{N_s}{N_p})V_p[/tex]
[tex]V_s=3V_p[/tex]...............(1)
The power dissipated through the secondary coil is :
[tex]P_s=\dfrac{V_s^2}{R}[/tex]
[tex]100\ W=\dfrac{V_s^2}{R}[/tex]
[tex]V_p^2=\dfrac{100R}{9}[/tex].............(2)
Let [tex]N_p'\ and\ N_s'[/tex] are the new number of turns in primary and secondary coil of the transformer such that,
[tex]\dfrac{N_p'}{N_s'}=\dfrac{1}{24}[/tex]
New voltage is :
[tex]V_s'=(\dfrac{N_s'}{N_p'})V_p'[/tex]
[tex]V_s'=24V_p[/tex]...............(3)
So, new power dissipated is [tex]P_s'[/tex]
[tex]P_s'=\dfrac{V_s'^2}{R}[/tex]
[tex]P_s'=\dfrac{(24V_p)^2}{R}[/tex]
[tex]P_s'=24^2\times \dfrac{(V_p)^2}{R}[/tex]
[tex]P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}[/tex]
[tex]P_s'=6400\ Watts[/tex]
So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.
