Neon sign require about 12,000 V for their operation. What should be the ratio of the number of loops in the secondary to the number of loops in the primary for a neon-sign transformer that operates off 120 V lines?
Nprimary/Nsecondary= Primaryvoltage/secondaryvoltage
solve for Nsec/Nprimary
To determine the ratio of the number of loops in the secondary (Nsecondary) to the number of loops in the primary (Nprimary) for a neon-sign transformer operating off 120V lines, we can use the formula:
Nprimary / Nsecondary = Primary voltage / Secondary voltage
Given that the neon sign requires about 12,000V (Secondary voltage) for operation, and the transformer is connected to 120V (Primary voltage) lines, we can substitute these values into the formula as follows:
Nprimary / Nsecondary = 120V / 12,000V
To get the ratio Nsecondary / Nprimary, we can take the reciprocal of both sides of the equation:
Nsecondary / Nprimary = 12,000V / 120V
Simplifying further:
Nsecondary / Nprimary = 100
Therefore, the ratio of the number of loops in the secondary (Nsecondary) to the number of loops in the primary (Nprimary) for this neon-sign transformer should be 100.