the transformer below has 58 primary turns, 102 secondary turns, and the output voltage is 126 v. what is…

the transformer below has 58 primary turns, 102 secondary turns, and the output voltage is 126 v. what is the input voltage of this transformer? 102 turns 58 turns secondary coil primary coil 126v output voltage soft iron core input voltage
Answer
Explanation:
Step1: Recall transformer voltage - turn ratio formula
For a transformer, $\frac{V_p}{V_s}=\frac{N_p}{N_s}$, where $V_p$ is the primary (input) voltage, $V_s$ is the secondary (output) voltage, $N_p$ is the number of primary turns, and $N_s$ is the number of secondary turns.
Step2: Rearrange the formula to solve for $V_p$
From $\frac{V_p}{V_s}=\frac{N_p}{N_s}$, we can get $V_p = V_s\times\frac{N_p}{N_s}$. Given $V_s = 126$ V, $N_p=58$, and $N_s = 102$. Substitute the values into the formula: $V_p=126\times\frac{58}{102}$. First, calculate $\frac{58}{102}=\frac{29}{51}$. Then $V_p = 126\times\frac{29}{51}=\frac{126\times29}{51}$. $126 = 2\times63$, $51=3\times17$, $126\times29=(2\times63)\times29$, $63 = 7\times9$. $126\times29=3654$, and $\frac{3654}{51}=71.65$ (rounded to two decimal places).
Answer:
$71.65$ V