301. Study the table and identify the row with suitable transformer notation.
| Row | Symbol | Meaning |
|---|
| P | \(N_p\) | Number of turns in the primary coil |
| Q | \(N_s\) | Primary current |
| R | \(V_p\) | Secondary voltage |
| S | \(I_s\) | Number of turns in the secondary coil |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row P
Explanation: The symbol \(N_p\) represents the number of turns in the primary coil. The symbol \(N_s\) represents the number of turns in the secondary coil, not primary current. Primary voltage is written as \(V_p\), while secondary voltage is written as \(V_s\). Secondary current is written as \(I_s\), not as a turns value. Correct notation is essential because transformer relations connect voltage, current, and turns ratio.
302. In an ideal transformer, the voltage ratio is related to the turns ratio by
ⓐ. \(\frac{V_s}{V_p}=\frac{N_p}{N_s}\)
ⓑ. \(\frac{V_s}{V_p}=N_sN_p\)
ⓒ. \(\frac{V_s}{V_p}=V_s+V_p\)
ⓓ. \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)
Correct Answer: \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)
Explanation: In an ideal transformer, the induced emf in a coil is proportional to the number of turns linked by the changing magnetic flux. Therefore, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns. This gives \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\). The relation assumes ideal coupling and negligible losses. Interchanging the turns ratio would reverse the prediction for step-up and step-down transformers.
303. When \(N_s\gt N_p\), a transformer is best described as a
ⓐ. step-up transformer
ⓑ. step-down transformer
ⓒ. pure resistor
ⓓ. rectifier
Correct Answer: step-up transformer
Explanation: If \(N_s\gt N_p\), then the secondary coil has more turns than the primary coil. From \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\), the secondary voltage becomes greater than the primary voltage. Such a transformer is called a step-up transformer. It raises voltage while reducing current in the ideal power-conservation picture. The term step-up refers to voltage, not to an increase in energy.
304. For an ideal transformer with \(N_p=500\), \(N_s=2500\), and \(V_p=220\,\text{V}\), the secondary voltage is
ⓐ. \(44\,\text{V}\)
ⓑ. \(1100\,\text{V}\)
ⓒ. \(220\,\text{V}\)
ⓓ. \(2500\,\text{V}\)
Correct Answer: \(1100\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(N_p=500\), \(N_s=2500\), and \(V_p=220\,\text{V}\).
\( \textbf{Required:} \) Secondary voltage \(V_s\).
\( \textbf{Transformer voltage relation:} \)
\[
\frac{V_s}{V_p}=\frac{N_s}{N_p}
\]
\( \textbf{Rearranging:} \)
\[
V_s=V_p\frac{N_s}{N_p}
\]
\( \textbf{Substitution:} \)
\[
V_s=220\left(\frac{2500}{500}\right)\,\text{V}
\]
\( \textbf{Turns ratio:} \)
\[
\frac{2500}{500}=5
\]
\( \textbf{Calculation:} \)
\[
V_s=220(5)=1100\,\text{V}
\]
\( \textbf{Final answer:} \) The ideal secondary voltage is \(1100\,\text{V}\).
305. If \(N_s\lt N_p\), the correct statement for ideal transformer operation is that it
ⓐ. steps down voltage and steps up current
ⓑ. steps up voltage and steps down current
ⓒ. steps down both voltage and current without changing power
ⓓ. works only with steady \(\text{DC}\)
Correct Answer: steps down voltage and steps up current
Explanation: If \(N_s\lt N_p\), the transformer is step-down because \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\) gives \(V_s\lt V_p\). In an ideal transformer, input power and output power are equal, so \(V_pI_p=V_sI_s\). If the secondary voltage is smaller, the secondary current is larger for the same ideal power transfer. This does not mean energy is created; voltage and current adjust inversely. The transformer still requires changing current, so steady \(\text{DC}\) is not suitable for ordinary transformer action.
306. In an ideal transformer, power conservation is expressed as
ⓐ. \(V_pV_s=I_pI_s\)
ⓑ. \(V_p+I_p=V_s+I_s\)
ⓒ. \(I_s=I_p\) for all turns ratios
ⓓ. \(V_pI_p=V_sI_s\)
Correct Answer: \(V_pI_p=V_sI_s\)
Explanation: An ideal transformer is assumed to have no energy losses. Therefore, the input power supplied to the primary equals the output power delivered by the secondary. Since electrical power is voltage multiplied by current, the ideal relation is \(V_pI_p=V_sI_s\). This relation explains why increasing voltage reduces current and decreasing voltage increases current. Equal power does not require equal voltage or equal current separately.
307. For an ideal transformer, the current ratio is
ⓐ. \(\frac{I_s}{I_p}=\frac{N_p}{N_s}\)
ⓑ. \(\frac{I_s}{I_p}=\frac{N_s}{N_p}\)
ⓒ. \(\frac{I_s}{I_p}=N_sN_p\)
ⓓ. \(\frac{I_s}{I_p}=\frac{V_s}{V_p}\)
Correct Answer: \(\frac{I_s}{I_p}=\frac{N_p}{N_s}\)
Explanation: In an ideal transformer, \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\). Also, power conservation gives \(V_pI_p=V_sI_s\). Rearranging the power relation gives \(\frac{I_s}{I_p}=\frac{V_p}{V_s}\). Substituting the voltage-turns relation gives \(\frac{I_s}{I_p}=\frac{N_p}{N_s}\). Current ratio is inverse to turns ratio, which is why a step-up transformer gives lower secondary current.
308. An ideal transformer steps \(240\,\text{V}\) down to \(24\,\text{V}\). If the primary current is \(0.5\,\text{A}\), the secondary current is
ⓐ. \(0.05\,\text{A}\)
ⓑ. \(0.5\,\text{A}\)
ⓒ. \(50\,\text{A}\)
ⓓ. \(5.0\,\text{A}\)
Correct Answer: \(5.0\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(V_p=240\,\text{V}\), \(V_s=24\,\text{V}\), and \(I_p=0.5\,\text{A}\).
\( \textbf{Required:} \) Secondary current \(I_s\).
\( \textbf{Ideal power relation:} \)
\[
V_pI_p=V_sI_s
\]
\( \textbf{Rearranging:} \)
\[
I_s=\frac{V_pI_p}{V_s}
\]
\( \textbf{Substitution:} \)
\[
I_s=\frac{(240)(0.5)}{24}\,\text{A}
\]
\( \textbf{Numerator:} \)
\[
240\times0.5=120
\]
\( \textbf{Calculation:} \)
\[
I_s=\frac{120}{24}=5.0\,\text{A}
\]
\( \textbf{Final answer:} \) The secondary current is \(5.0\,\text{A}\).
309. In ideal transformer operation, \(V_s=4V_p\). The relation between secondary and primary currents is
ⓐ. \(I_s=4I_p\)
ⓑ. \(I_s=2I_p\)
ⓒ. \(I_s=\frac{I_p}{4}\)
ⓓ. \(I_s=\frac{I_p}{2}\)
Correct Answer: \(I_s=\frac{I_p}{4}\)
Explanation: In an ideal transformer, input power equals output power, so \(V_pI_p=V_sI_s\). If \(V_s=4V_p\), then \(V_pI_p=4V_pI_s\). Cancelling \(V_p\) gives \(I_p=4I_s\). Therefore, \(I_s=\frac{I_p}{4}\). A voltage increase is accompanied by a current decrease so that power is not created.
310. Read the passage and answer the question.
A transformer has \(N_p=1000\) and \(N_s=200\). The primary is connected to a \(230\,\text{V}\) alternating supply. Losses are neglected.
The secondary voltage is closest to
ⓐ. \(23\,\text{V}\)
ⓑ. \(230\,\text{V}\)
ⓒ. \(46\,\text{V}\)
ⓓ. \(1150\,\text{V}\)
Correct Answer: \(46\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(N_p=1000\), \(N_s=200\), and \(V_p=230\,\text{V}\).
\( \textbf{Transformer relation:} \)
\[
\frac{V_s}{V_p}=\frac{N_s}{N_p}
\]
\( \textbf{Rearranging:} \)
\[
V_s=V_p\frac{N_s}{N_p}
\]
\( \textbf{Substitution:} \)
\[
V_s=230\left(\frac{200}{1000}\right)\,\text{V}
\]
\( \textbf{Turns ratio:} \)
\[
\frac{200}{1000}=0.2
\]
\( \textbf{Calculation:} \)
\[
V_s=230(0.2)=46\,\text{V}
\]
\( \textbf{Final answer:} \) The secondary voltage is \(46\,\text{V}\).
311. A student says: “A step-up transformer increases voltage, so it also increases power.” The statement is unsuitable because an ideal step-up transformer
ⓐ. increases voltage while decreasing current
ⓑ. increases both voltage and current without any limit
ⓒ. works only when no secondary coil is present
ⓓ. has zero secondary voltage
Correct Answer: increases voltage while decreasing current
Explanation: A step-up transformer has \(N_s\gt N_p\), so it increases secondary voltage. In the ideal case, however, input power equals output power. Therefore, the increase in voltage is accompanied by a decrease in current. The relation \(V_pI_p=V_sI_s\) prevents ideal power gain. A transformer changes voltage and current levels; it does not create energy.
312. For an ideal transformer with \(N_p=400\), \(N_s=1600\), and \(I_s=2\,\text{A}\), the primary current is
ⓐ. \(0.5\,\text{A}\)
ⓑ. \(2\,\text{A}\)
ⓒ. \(8\,\text{A}\)
ⓓ. \(4\,\text{A}\)
Correct Answer: \(8\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(N_p=400\), \(N_s=1600\), and \(I_s=2\,\text{A}\).
\( \textbf{Required:} \) Primary current \(I_p\).
\( \textbf{Ideal current-turns relation:} \)
\[
\frac{I_s}{I_p}=\frac{N_p}{N_s}
\]
\( \textbf{Substitution:} \)
\[
\frac{2}{I_p}=\frac{400}{1600}
\]
\( \textbf{Turns ratio simplification:} \)
\[
\frac{400}{1600}=\frac{1}{4}
\]
\( \textbf{Solving:} \)
\[
\frac{2}{I_p}=\frac{1}{4}
\]
\[
I_p=8\,\text{A}
\]
\( \textbf{Final answer:} \) The primary current is \(8\,\text{A}\).
313. A changing primary current in a transformer produces a secondary emf of the same frequency mainly because
ⓐ. the same changing flux links both coils
ⓑ. the secondary coil is connected directly to the primary wire
ⓒ. the iron core acts as a battery
ⓓ. the number of secondary turns must equal the number of primary turns
Correct Answer: the same changing flux links both coils
Explanation: The primary alternating current produces a changing magnetic flux in the core. This flux links the secondary coil and changes with the same frequency as the primary source in the ideal picture. The induced emf in the secondary therefore has the same frequency as the changing flux. The transformer changes voltage and current according to turns ratio, not the frequency of the supply. Direct electrical contact between primary and secondary is not required for ordinary transformer action.
314. The set of suitable statements for an ideal transformer is:
I. \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)
II. \(V_pI_p=V_sI_s\)
III. A step-up transformer has \(N_s\lt N_p\)
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is true because ideal transformer voltage ratio equals turns ratio. Statement II is true because ideal operation assumes no power loss, so input power equals output power. Statement III is false because a step-up transformer has more turns in the secondary than in the primary, so \(N_s\gt N_p\). If \(N_s\lt N_p\), the transformer is step-down. The transformer relations must be interpreted with both turns ratio and power conservation together.
315. The efficiency of a transformer is defined as
ⓐ. \(\eta=\frac{\text{input power}}{\text{output power}}\times100\%\)
ⓑ. \(\eta=\frac{V_s}{I_s}\times100\%\)
ⓒ. \(\eta=\frac{N_p}{N_s}\times100\%\)
ⓓ. \(\eta=\frac{\text{output power}}{\text{input power}}\times100\%\)
Correct Answer: \(\eta=\frac{\text{output power}}{\text{input power}}\times100\%\)
Explanation: Transformer efficiency compares useful output power with the input power supplied to the primary. It is written as \(\eta=\frac{P_{\text{out}}}{P_{\text{in}}}\times100\%\). An ideal transformer would have \(\eta=100\%\), but practical transformers have losses. The ratio \(\frac{V_s}{I_s}\) does not represent efficiency because voltage divided by current gives an impedance-like quantity. The turns ratio decides voltage transformation, while efficiency measures how much input power reaches the load.
316. A transformer takes \(500\,\text{W}\) from the primary supply and delivers \(450\,\text{W}\) to the load. Its efficiency is
ⓐ. \(50\%\)
ⓑ. \(90\%\)
ⓒ. \(80\%\)
ⓓ. \(95\%\)
Correct Answer: \(90\%\)
Explanation: \( \textbf{Given:} \) Input power \(P_{\text{in}}=500\,\text{W}\), output power \(P_{\text{out}}=450\,\text{W}\).
\( \textbf{Required:} \) Efficiency \(\eta\).
\( \textbf{Efficiency relation:} \)
\[
\eta=\frac{P_{\text{out}}}{P_{\text{in}}}\times100\%
\]
\( \textbf{Substitution:} \)
\[
\eta=\frac{450}{500}\times100\%
\]
\( \textbf{Power ratio:} \)
\[
\frac{450}{500}=0.90
\]
\( \textbf{Calculation:} \)
\[
\eta=0.90\times100\%=90\%
\]
\( \textbf{Loss interpretation:} \) The remaining \(50\,\text{W}\) is lost in practical transformer losses.
\( \textbf{Final answer:} \) The transformer efficiency is \(90\%\).
317. Copper loss in a transformer is mainly due to
ⓐ. absence of turns in the secondary coil
ⓑ. steady magnetic flux only
ⓒ. capacitance between the plates of a capacitor
ⓓ. winding resistance in the coils
Correct Answer: winding resistance in the coils
Explanation: Transformer windings are made of conducting wire, and real wires have resistance. When current flows through these windings, heat is produced according to \(I^2R\). This heating loss is called copper loss because copper is commonly used for windings. It occurs in both primary and secondary coils. It is reduced by using low-resistance windings of suitable thickness.
318. The loss that is reduced by laminating the iron core of a transformer is mainly
ⓐ. copper loss
ⓑ. eddy current loss
ⓒ. loss due to zero secondary turns
ⓓ. loss due to absence of alternating flux
Correct Answer: eddy current loss
Explanation: A changing magnetic flux in the transformer core can induce circulating currents inside the core material. These currents are called eddy currents, and they cause heating loss. Laminating the core divides it into thin insulated sheets, which breaks the paths of large circulating currents. This greatly reduces eddy current loss. Copper loss is instead reduced by lowering the resistance of the windings.
319. A soft iron core is preferred in a transformer because it
ⓐ. has high hysteresis loss and poor magnetic response
ⓑ. prevents any magnetic flux from linking the secondary
ⓒ. reduces hysteresis loss and improves magnetic coupling
ⓓ. makes the transformer suitable for steady \(\text{DC}\) only
Correct Answer: reduces hysteresis loss and improves magnetic coupling
Explanation: The transformer core undergoes repeated magnetisation and demagnetisation when supplied with \(\text{AC}\). A soft iron core has a narrow hysteresis loop, so less energy is lost in each magnetic cycle. It also provides a low-reluctance path for magnetic flux, improving coupling between primary and secondary coils. This helps more of the changing flux link both windings. The core material is chosen for magnetic efficiency, not for blocking flux.
320. Study the table and identify the suitable row about transformer losses and their reduction.
| Row | Loss | Suitable reduction method |
|---|
| P | Eddy current loss | Use a laminated core |
| Q | Copper loss | Use very high-resistance winding wire |
| R | Hysteresis loss | Use a hard magnetic material with wide loop |
| S | Flux leakage | Remove the iron core completely |
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Eddy current loss is reduced by using a laminated core, because laminations restrict circulating currents in the core. Copper loss is reduced by using low-resistance winding wire, not high-resistance wire. Hysteresis loss is reduced by using a soft magnetic material with low hysteresis loss. Flux leakage is reduced by good core design and tight magnetic coupling, not by removing the core. The correct row pairs the loss with the construction feature that directly reduces it.
