Question

An ideal efficient transformer has a primary power input of 10kW. The secondary current when the transformer is on load is 25A. If the primary: secondary turns ratio is 8:1, then the potential difference applied to the primary coil is –

& \text{A) }\dfrac{{{10}^{4}}\times {{8}^{2}}}{25}V \\\ & \text{B) }\dfrac{{{10}^{4}}\times 8}{25}V \\\ & \text{C) }\dfrac{{{10}^{4}}}{25\times 8}V \\\ & \text{D) }\dfrac{{{10}^{4}}}{25\times {{8}^{2}}} \\\ \end{aligned}$$

Answer 1

We are given transformer parameters from which we have to find the potential difference provided to the primary coil to induce the emf in the secondary coil. We can use the transformer equations and easily find the solution to the problem.

Complete step by step answer:
We know that the transformer works on the principle of mutual induction. The primary coil is fed with a voltage or potential difference which induces an emf in the secondary coil. From the mutual inductance we know that in a transformer the voltage and current in the primary coil is related to the voltage and current in the secondary coil as –

& \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{I}_{2}}}{{{I}_{1}}} \\\ & {{V}_{1}}{{\text{I}}_{1}}\text{= }{{V}_{2}}{{I}_{2}} \\\ \end{aligned}$$ Where, $${{V}_{1}}\text{ and }{{V}_{2}}$$ is the voltage in the primary and secondary coils respectively, $${{I}_{1}}\text{ and }{{I}_{2}}$$ is the current in the primary and secondary coils respectively. We also know that the turns in both the coils are also related to the voltage and current as – $$\dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{I}_{2}}}{{{I}_{1}}}$$ We know that the power dissipated in the coil can be given as – $$P=VI$$ We can relate the coils as – $$\begin{aligned} & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{{{I}_{2}}}{{{I}_{1}}} \\\ & \Rightarrow \dfrac{8}{1}=\dfrac{25}{{{I}_{1}}} \\\ & \Rightarrow {{I}_{1}}=\dfrac{25}{8}A \\\ \end{aligned}$$ Using the relation with the power and potential difference for the primary coil, we get – $$\begin{aligned} & {{P}_{1}}={{V}_{1}}{{I}_{1}} \\\ & \Rightarrow 10kW={{V}_{1}}\dfrac{25}{8}A \\\ & \Rightarrow {{V}_{1}}=\dfrac{10\times {{10}^{3}}\times 8}{25} \\\ & \therefore {{V}_{1}}=\dfrac{{{10}^{4}}\times 8}{25}V \\\ \end{aligned}$$ The potential difference of the primary coil is $$\dfrac{{{10}^{4}}\times 8}{25}V$$ **The correct answer is option B.** **Note:** The mutual inductance between the coils is dependent on the voltage, current and the number of turns in both the primary and secondary coils. The inductance is increased by using a core inside the coil which has a higher magnetic permeability than air.