An ideal transformer has 255 loops in its primary coil and 125 loops in its secondary coil.If the primary voltage is 120 V_rms and the device connected to the secondary uses 100 W of power,what is the current in the secondary loop? A)0.41 A B)0.59 A C)1.7 A D)59 A E)240 A

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Quizplus · Accepted Answer

The power used in the secondary circuit is given as 100 W, and power (P) is the product of voltage (V) and current (I), i.e., $P = VI$. To find the secondary voltage ($V_s$), we use the transformer equation relating primary and secondary voltages and the number of loops (turns) in their coils: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$, where $V_p$ and $V_s$ are the primary and secondary voltages, and $N_p$ and $N_s$ are the number of loops in the primary and secondary coils, respectively. Given $V_p = 120$ V, $N_p = 255$, and $N_s = 125$, we can solve for $V_s$:$$\frac{120}{V_s} = \frac{255}{125} \Rightarrow V_s = \frac{120 \times 125}{255} = 58.82 \text{ V (approximately)}$$Now, using $P = VI$ for the secondary circuit, where $P = 100$ W and $V_s = 58.82$ V, we solve for the secondary current ($I_s$):$$100 = 58.82 \times I_s \Rightarrow I_s = \frac{100}{58.82} = 1.7 \text{ A (approximately)}$$Therefore, the current in the secondary loop is approximately 1.7 A.

An Ideal Transformer Has 255 Loops in Its Primary Coil