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Chapter 14: Problem 2034

A transformer of efficiency \(90 \%\) draws an input power of \(4 \mathrm{~kW}\). An electrical appliance connected across the secondary draws a current of $6 \mathrm{~A}$. The impedance of device is ....... (a) \(60 \Omega\) (b) \(50 \Omega\) (c) \(80 \Omega\) (d) \(100 \Omega\)

Short Answer

Expert verified
The impedance of the device is \(100 \Omega\), which corresponds to option (d).

Step by step solution

01

Calculate the output power of the transformer

We are given the efficiency of the transformer as 90% and the input power as 4 kW. The efficiency of the transformer is defined as the ratio of the output power to the input power. Let's denote the output power as P_out (in kW). Therefore, Efficiency = \(\frac{P_{out}}{P_{in}}\) We are given Efficiency = 0.9 and P_in = 4 kW. We will solve for P_out: 0.9 = \(\frac{P_{out}}{4}\) P_out = 0.9 * 4 = 3.6 kW
02

Convert the output power to Watts

Since we are given the current drawn by the electrical appliance in Amperes, it is convenient to have the output power in Watts. To convert the output power from kilowatts to watts, we simply multiply by 1000: P_out (W) = 3.6 * 1000 = 3600 W
03

Calculate the impedance of the device using Ohm's law

Ohm's law relates the voltage V, current I, and impedance (resistance) R as: V = I * R In this case, we want to find the impedance (R) of the device. We also know that the power P_out equals the product of current I and voltage V across the device: P_out = V * I Since we have already found the output power and we are given the current I, we can now find the voltage V across the device: V = \(\frac{P_{out}}{I}\) V = \(\frac{3600}{6}\) = 600 V Now, using Ohm's law, we can find the impedance R: R = \(\frac{V}{I}\) R = \(\frac{600}{6}\) = 100 Ω Therefore, the impedance of the device is 100 Ω, which corresponds to option (d).

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