Sure! To solve for the electromotive force (e.m.f) in the secondary coil of a transformer, we can use the transformer equation, which relates the voltages and the number of turns in the primary and secondary coils. The equation is as follows:
[tex]\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \][/tex]
where:
- [tex]\( V_s \)[/tex] is the voltage (e.m.f) in the secondary coil
- [tex]\( V_p \)[/tex] is the voltage (e.m.f) in the primary coil
- [tex]\( N_s \)[/tex] is the number of turns in the secondary coil
- [tex]\( N_p \)[/tex] is the number of turns in the primary coil
Given:
- [tex]\( V_p = 240 \)[/tex] volts
- [tex]\( N_p = 1000 \)[/tex] turns
- [tex]\( N_s = 500 \)[/tex] turns
We need to find [tex]\( V_s \)[/tex], the voltage in the secondary coil. Rearrange the formula to solve for [tex]\( V_s \)[/tex]:
[tex]\[ V_s = V_p \times \frac{N_s}{N_p} \][/tex]
Substitute the given values into the equation:
[tex]\[ V_s = 240 \, \text{V} \times \frac{500}{1000} \][/tex]
Simplify the fraction:
[tex]\[ \frac{500}{1000} = 0.5 \][/tex]
So,
[tex]\[ V_s = 240 \, \text{V} \times 0.5 \][/tex]
Perform the multiplication:
[tex]\[ V_s = 120 \, \text{V} \][/tex]
Therefore, the e.m.f in the secondary coil is [tex]\( 120 \)[/tex] volts.
