67.8 turns needed by the secondary coil to run the bulb.
Explanation:
We know that,
[tex]\text { Electric power }(p)=\frac{V^{2}}{R}[/tex]
[tex]\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}[/tex]
[tex]\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}[/tex]
For calculating number of turns
[tex]\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}[/tex]
Given that,
[tex]80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }[/tex]
[tex]\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}[/tex]
[tex]\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .[/tex]
We need to find the number of turns in the secondary winding [tex]\left(N_{S}\right)[/tex] to run the bulb at 120W [tex]\left(P_{2}\right)[/tex]
Firstly find the secondary voltage in the transformer use, [tex]\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}[/tex]
[tex]\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}[/tex]
[tex]V_{2}^{2}=\frac{120^{2} \times 120}{80}[/tex]
[tex]V_{2}^{2}=\frac{1728000}{80}[/tex]
[tex]V_{2}^{2}=21600[/tex]
[tex]V_{2}=\sqrt{21600}[/tex]
[tex]V_{2}=146.9 \mathrm{V}=V_{S}[/tex]
Now, finding the number of turns in secondary coil. Use, [tex]\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}[/tex]
[tex]\frac{30}{N_{S}}=\frac{65}{146.9}[/tex]
[tex]N_{S}=\frac{30 \times 146.9}{65}[/tex]
[tex]N_{S}=\frac{4407}{65}[/tex][tex]N_{S}=67.8[/tex]
The number of turns in the secondary winding are 67.8 turns.
