Transformer Questions in English

(D) For maximum power transfer,the impedance of the source must match the impedance of the load as seen through the transformer.
Let the primary circuit have impedance $Z_1$ and the secondary circuit have impedance $Z_2$.
The transformer transforms the load impedance $Z_2$ to the primary side as $Z_2' = (N_1/N_2)^2 Z_2$.
For maximum power transfer,the primary impedance $Z_1$ must equal the reflected load impedance $Z_2'$.
So,$Z_1 = (N_1/N_2)^2 Z_2$.
Rearranging this,we get $(N_1/N_2)^2 = Z_1/Z_2$.
Taking the square root on both sides,we get $N_1/N_2 = \sqrt{Z_1/Z_2}$.

(D) The turn ratio of the transformer is given by $\frac{N_2}{N_1} = \frac{8}{1}$.
The relation between primary voltage $(V_1)$ and secondary voltage $(V_2)$ is $\frac{V_2}{V_1} = \frac{N_2}{N_1}$.
Given $V_1 = 120\, V$,we calculate $V_2$ as:
$V_2 = V_1 \times \frac{N_2}{N_1} = 120 \times 8 = 960\, V$.
The current in the secondary coil $(I_2)$ is determined by Ohm's Law using the load resistance $(R_L = 10^4\ \Omega)$:
$I_2 = \frac{V_2}{R_L} = \frac{960}{10^4} = 0.096\, A$.
Converting to milliamperes:
$I_2 = 0.096 \times 1000\, mA = 96\, mA$.

(A) The turn ratio of the transformer is given as $k = \frac{N_s}{N_p} = 5$.
The $RMS$ voltage of the primary coil is $V_{p(rms)} = 100 \, V$.
For an ideal transformer,the relation between the secondary voltage $(V_s)$ and primary voltage $(V_p)$ is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Therefore,the $RMS$ voltage of the secondary coil is $V_{s(rms)} = 5 \times 100 \, V = 500 \, V$.
The peak value of the voltage $(V_0)$ is related to the $RMS$ value by the formula $V_0 = \sqrt{2} \times V_{rms}$.
Thus,the peak value of the voltage in the secondary winding is $V_{s(peak)} = \sqrt{2} \times 500 = 500 \sqrt{2} \, V$.

(C) The efficiency $\eta$ of a transformer is defined as the ratio of output power $(P_o)$ to input power $(P_i)$, expressed as a percentage: $\eta = \frac{P_o}{P_i} \times 100$.
The input power $P_i$ is calculated using the voltage and current from the mains: $P_i = V_{in} \times I_{in} = 240\, V \times 0.7\, A = 168\, W$.
The output power $P_o$ is given as $140\, W$.
Substituting these values into the efficiency formula:
$\eta = \frac{140}{168} \times 100 = \frac{5}{6} \times 100 \approx 83.33\%$.
Therefore, the efficiency of the transformer is approximately $83.3\%$.

(C) Given: Primary turns $N_1 = 100$, Primary current $I_1 = 8$ $A$, Input power $P_{in} = 1000$ $W$, Secondary voltage $V_2 = 500$ $V$.
First, calculate the primary voltage $V_1$ using $P_{in} = V_1 I_1$:
$1000 = V_1 \times 8 \Rightarrow V_1 = 125$ $V$.
For an ideal transformer, the ratio of turns is equal to the ratio of voltages:
$\frac{N_2}{N_1} = \frac{V_2}{V_1}$
$\frac{N_2}{100} = \frac{500}{125}$
$\frac{N_2}{100} = 4$
$N_2 = 400$ turns.

(D) transformer works on the principle of electromagnetic induction,which requires a time-varying magnetic flux ($AC$ current).
$1$. The input provided is a cell,which produces a constant $D.C.$ voltage.
$2$. $A$ $D.C.$ voltage does not produce a changing magnetic flux in the primary coil.
$3$. Since there is no change in magnetic flux,no induced $e.m.f.$ is generated in the secondary coil according to Faraday's Law of Induction.
$4$. Therefore,the secondary voltage developed is $0 \, V$.

(A) For an ideal transformer,the power input equals the power output,which implies $V_{p} i_{p} = V_{s} i_{s}$.
Since the voltage ratio is proportional to the turns ratio,we have $\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}}$.
Given: $N_{p} = 500$,$N_{s} = 2000$,and $i_{p} = 48 \ A$.
Substituting the values into the relation $\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}}$:
$\frac{2000}{500} = \frac{48}{i_{s}}$
$4 = \frac{48}{i_{s}}$
$i_{s} = \frac{48}{4} = 12 \ A$.

(C) Given: Primary voltage $V_{P} = 2300\,V$,Secondary voltage $V_{S} = 230\,V$,Primary current $I_{P} = 5\,A$,Efficiency $\eta = 90\% = 0.9$.
The efficiency of a transformer is defined as the ratio of output power $(P_{S})$ to input power $(P_{P})$:
$\eta = \frac{P_{S}}{P_{P}} \Rightarrow P_{S} = \eta \times P_{P}$.
Since power $P = V \times I$,we can write:
$V_{S} \times I_{S} = 0.9 \times (V_{P} \times I_{P})$.
Substituting the given values:
$230 \times I_{S} = 0.9 \times 2300 \times 5$.
Solving for $I_{S}$:
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5 = 45\,A$.
Therefore,the output current is $45\,A$.

(B) The efficiency $\eta$ of a transformer is defined as the ratio of output power $P_s$ to input power $P_p$,i.e.,$\eta = \frac{P_s}{P_p}$.
Given: Input voltage $V_p = 2300 \, V$,Input current $I_p = 5 \, A$,Output voltage $V_s = 230 \, V$,and Efficiency $\eta = 90 \% = 0.9$.
Input power $P_p = V_p \times I_p = 2300 \, V \times 5 \, A = 11500 \, W$.
Output power $P_s = \eta \times P_p = 0.9 \times 11500 \, W = 10350 \, W$.
Since $P_s = V_s \times I_s$,the output current $I_s = \frac{P_s}{V_s} = \frac{10350 \, W}{230 \, V} = 45 \, A$.

(A) Given: Number of turns in primary $N_{p} = 300$,Number of turns in secondary $N_{s} = 150$,Output power $P_{s} = 2.2\, kW = 2200\, W$,Secondary current $I_{s} = 10\, A$.
First,calculate the secondary voltage $V_{s}$ using $P_{s} = V_{s} I_{s}$:
$2200 = V_{s} \times 10 \Rightarrow V_{s} = 220\, V$.
Using the transformer ratio $\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$:
$\frac{V_{p}}{220} = \frac{300}{150} = 2 \Rightarrow V_{p} = 440\, V$.
Assuming an ideal transformer,input power $P_{p} = P_{s} = 2200\, W$:
$P_{p} = V_{p} I_{p} \Rightarrow 2200 = 440 \times I_{p} \Rightarrow I_{p} = \frac{2200}{440} = 5\, A$.
Thus,the input voltage is $440\, V$ and the input current is $5\, A$.

(D) Given:
Primary voltage, $V_{P} = 11000 \, V$
Secondary voltage, $V_{S} = 220 \, V$
Number of primary turns, $N_{P} = 6000$
Efficiency, $\eta = 60 \% = 0.6$
Output power, $P_{O} = 9 \, kW$
Efficiency is defined as the ratio of output power to input power:
$\eta = \frac{P_{O}}{P_{i}}$
Rearranging the formula to find the input power $(P_{i})$:
$P_{i} = \frac{P_{O}}{\eta}$
Substituting the given values:
$P_{i} = \frac{9 \, kW}{0.6}$
$P_{i} = 15 \, kW$
Therefore, the input power is $15 \, kW$.

(A) transformer operates on the principle of electromagnetic induction, which requires a changing magnetic flux to induce an electromotive force $(EMF)$ in the secondary coil.
Since a battery provides a direct current $(DC)$ supply, the current flowing through the primary coil is constant.
A constant current produces a constant magnetic flux, which does not change with time.
According to Faraday's law of induction, the induced $EMF$ is proportional to the rate of change of magnetic flux $(e = -d\Phi/dt)$.
Since the flux is constant, $d\Phi/dt = 0$, and therefore, the induced secondary voltage is $0\, V$.

(D) Given,the turn ratio $\frac{N_s}{N_p} = 8:1 = 8$.
Input voltage $V_p = 120\,V$.
For an ideal transformer,the ratio of voltages is equal to the turn ratio:
$\frac{V_s}{V_p} = \frac{N_s}{N_p} = 8$.
Therefore,the secondary voltage $V_s = 8 \times V_p = 8 \times 120\,V = 960\,V$.
The load resistance $R_s = 10^4\,\Omega$.
Using Ohm's law,the current in the secondary coil $I_s = \frac{V_s}{R_s}$.
$I_s = \frac{960}{10^4} = 960 \times 10^{-4} = 96 \times 10^{-3}\,A$.
$I_s = 96\,mA$.

(C) For an ideal transformer,the power input equals the power output,assuming no energy losses.
The relationship is given by $V_p I_p = V_s I_s$,where $V_p$ and $I_p$ are the primary voltage and current,and $V_s$ and $I_s$ are the secondary voltage and current.
Given:
Primary voltage $V_p = 4000 \, V$
Primary current $I_p = 100 \, A$
Secondary voltage $V_s = 240,000 \, V$
Substituting the values into the equation:
$4000 \times 100 = 240,000 \times I_s$
$400,000 = 240,000 \times I_s$
$I_s = \frac{400,000}{240,000}$
$I_s = \frac{40}{24} = \frac{5}{3} \approx 1.67 \, A$.

(A) For a $100\%$ efficient transformer, the input power equals the output power.
$P_{in} = P_{out}$
$V_{p} I_{p} = V_{s} I_{s}$
We know that for an ideal transformer, the ratio of currents is inversely proportional to the ratio of turns:
$\frac{I_{p}}{I_{s}} = \frac{N_{s}}{N_{p}}$
Given:
$N_{p} = 100$
$N_{s} = 25$
$I_{s} = 4 \text{ A}$
Substituting the values:
$\frac{I_{p}}{4} = \frac{25}{100}$
$\frac{I_{p}}{4} = \frac{1}{4}$
$I_{p} = 1 \text{ A}$
Therefore, the current in the primary coil is $1 \text{ A}$.

(D) The efficiency of a transformer is defined as the ratio of output power to input power.
Efficiency $(\eta) = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100 = \frac{V_s I_s}{V_p I_p} \times 100$
Given:
Primary voltage $V_p = 220\, V$,Primary current $I_p = 5\, A$
Secondary voltage $V_s = 11\, V$,Secondary current $I_s = 90\, A$
Input power $P_{\text{input}} = V_p \times I_p = 220 \times 5 = 1100\, W$
Output power $P_{\text{output}} = V_s \times I_s = 11 \times 90 = 990\, W$
Efficiency $\eta = \frac{990}{1100} \times 100 = 0.9 \times 100 = 90\, \%$

(B) For an ideal transformer,the input power is equal to the output power.
Input power $P_{in} = E_{P} I_{P}$
Output power $P_{out} = E_{S} I_{S} = 24 \, W$
Since $P_{in} = P_{out}$,we have $E_{P} I_{P} = 24 \, W$.
Given $E_{P} = 240 \, V$,we can calculate the primary current $I_{P}$ as:
$I_{P} = \frac{24 \, W}{240 \, V} = 0.1 \, A$.

(C) For an ideal transformer with $100\%$ efficiency, the input power equals the output power: $P_{in} = P_{out}$.
Since $P = V \cdot I$, we have $V_p \cdot I_p = V_s \cdot I_s$.
The transformer ratio is given by $\frac{V_p}{V_s} = \frac{n_p}{n_s} = \frac{I_s}{I_p}$.
Given $\frac{n_p}{n_s} = 20:1$ and $I_s = 80\,A$.
Substituting the values: $\frac{I_s}{I_p} = \frac{n_p}{n_s} \implies \frac{80}{I_p} = \frac{20}{1}$.
Solving for $I_p$: $I_p = \frac{80}{20} = 4\,A$.

(B) Input voltage,$V_1 = 2300\; V$.
Number of turns in the primary coil,$n_1 = 4000$.
Output voltage,$V_2 = 230\; V$.
Let the number of turns in the secondary coil be $n_2$.
The relationship between voltage and the number of turns in a transformer is given by the formula:
$\frac{V_1}{V_2} = \frac{n_1}{n_2}$
Substituting the given values:
$\frac{2300}{230} = \frac{4000}{n_2}$
$10 = \frac{4000}{n_2}$
$n_2 = \frac{4000}{10} = 400$.
Thus,the number of turns in the secondary coil should be $400$.

(N/A) Total electric power required,$P = 800 \; kW = 800 \times 10^3 \; W$.
Supply voltage at the town,$V = 220 \; V$.
Voltage at which the electric plant generates power,$V' = 440 \; V$.
Distance between the town and the power generating station,$d = 15 \; km$.
Resistance of the two-wire line $= 0.5 \; \Omega/km$.
Total resistance of the wires,$R = (15 + 15) \times 0.5 = 15 \; \Omega$.
$A$ step-down transformer of rating $4000-220 \; V$ is used in the sub-station.
Input voltage to the transformer,$V_1 = 4000 \; V$.
The current in the transmission line is $I = \frac{P}{V_1} = \frac{800 \times 10^3}{4000} = 200 \; A$.
$(a)$ Line power loss $= I^2 R = (200)^2 \times 15 = 40000 \times 15 = 600,000 \; W = 600 \; kW$.
$(b)$ Total power supplied by the plant $= \text{Power required} + \text{Power loss} = 800 \; kW + 600 \; kW = 1400 \; kW$.
$(c)$ Voltage drop in the power line $= I \times R = 200 \times 15 = 3000 \; V$.
Total voltage transmitted from the plant $= \text{Voltage at transformer input} + \text{Voltage drop} = 4000 \; V + 3000 \; V = 7000 \; V$.
Since the plant generates at $440 \; V$,the step-up transformer rating is $440 \; V - 7000 \; V$.

(A) Given: Power $P = 800 \; kW = 8 \times 10^5 \; W$,Voltage at town $V_1 = 40,000 \; V$,Distance $d = 15 \; km$,Resistance per unit length $r = 0.5 \; \Omega/km$.
Total resistance $R = (15 + 15) \times 0.5 = 15 \; \Omega$.
Current in the line $I = P / V_1 = (800 \times 10^3) / 40,000 = 20 \; A$.
$(a)$ Line power loss $P_{loss} = I^2 R = (20)^2 \times 15 = 400 \times 15 = 6,000 \; W = 6 \; kW$.
$(b)$ Total power supplied by the plant $P_{total} = P + P_{loss} = 800 \; kW + 6 \; kW = 806 \; kW$.
$(c)$ Voltage drop in the line $V_{drop} = I \times R = 20 \times 15 = 300 \; V$. Voltage at the output of the step-up transformer $V_{out} = V_1 + V_{drop} = 40,000 + 300 = 40,300 \; V$. Since the plant generates at $440 \; V$,the step-up transformer is $440 \; V - 40,300 \; V$.
High-voltage transmission is preferred because power loss $P_{loss} = I^2 R = (P/V)^2 R$ is inversely proportional to the square of the transmission voltage $V^2$. Increasing $V$ significantly reduces $I^2 R$ losses.

(N/A) transformer is used to transmit voltage over large distances.
It is used because it can step up the voltage to a very high level for transmission,which significantly reduces the current flowing through the transmission lines.
Since power loss due to heating is given by $P_{loss} = I^2 R$,reducing the current $I$ minimizes energy loss during transmission over long distances.

(N/A) transformer is an electrical device that increases or decreases the $AC$ voltage.
Transformers that increase the voltage are called step-up transformers,while transformers that decrease the voltage are called step-down transformers.
Principle: $A$ transformer works on the principle of mutual electromagnetic induction.
Construction: $A$ transformer consists of two sets of coils insulated from each other.
As shown in figure $(a)$ and $(b)$,the coils are wound on a soft-iron core,either one on top of the other or on separate limbs of the core. One of the coils,called the primary coil,has $N_{p}$ turns.
The other coil is called the secondary coil and has $N_{s}$ turns. The primary coil is the input coil,and the secondary coil is the output coil of the transformer.
The coils are wound in two ways:
$(1)$ Core type: Coils are wound on a soft-iron core on separate limbs of the core as shown in figure $(b)$.
$(2)$ Shell type: Coils are wound on a soft-iron core,either one on top of the other as shown in figure $(a)$.

(N/A) When an alternating voltage is applied to the primary coil,the resulting current produces an alternating magnetic flux which links the secondary coil and induces an electromotive force (emf) in it.
The value of the induced emf depends on the number of turns in the secondary coil.
Let us consider an ideal transformer in which the primary coil has negligible resistance and all the magnetic flux in the core links both the primary and secondary windings.
If no load is connected to the secondary coil and $N_{p}$ and $N_{s}$ are the number of turns in the primary and secondary coils respectively,then the induced emf in the primary coil is:
$\varepsilon_{p} = -N_{p} \frac{d \phi}{d t} \quad \dots (1)$
And the induced emf in the secondary coil is:
$\varepsilon_{s} = -N_{s} \frac{d \phi}{d t} \quad \dots (2)$
Since $\varepsilon_{p} = V_{p}$ and $\varepsilon_{s} = V_{s}$,where $V_{p}$ and $V_{s}$ are the voltages across the primary and secondary coils respectively,equations $(1)$ and $(2)$ can be written as:
$V_{p} = -N_{p} \frac{d \phi}{d t}$ and $V_{s} = -N_{s} \frac{d \phi}{d t}$
Therefore,$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$.
The above relation is based on three assumptions:
$(1)$ The primary resistance and current are small.
$(2)$ The same magnetic flux links both the primary and secondary coils as very little flux escapes from the core.
$(3)$ The secondary current is small.
The ratio of the number of turns in the secondary coil $N_{s}$ to the number of turns in the primary coil $N_{p}$ is called the transformation ratio $\lambda$.
If the secondary coil has a greater number of turns than the primary $(N_{s} > N_{p})$,the voltage is stepped up. This is called a step-up transformer.
If the secondary coil has fewer turns than the primary $(N_{s} < N_{p})$,then $V_{s} < V_{p}$. This is called a step-down transformer,where the voltage is reduced.

(N/A) For an ideal transformer,the efficiency is $100 \%$,meaning there are no energy losses. Therefore,the input power is equal to the output power.
Since power $P = VI$,we have:
$\text{Input power} = \text{Output power}$
$I_p V_p = I_s V_s$
$\frac{I_p}{I_s} = \frac{V_s}{V_p} \quad \dots (1)$
For an ideal transformer,the ratio of voltages is equal to the ratio of the number of turns in the coils:
$\frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \dots (2)$
Combining equations $(1)$ and $(2)$,we get:
$\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$
This equation shows that the ratio of currents is inversely proportional to the ratio of voltages and turns.

(C) For an ideal transformer,the relationship is given by $\frac{I_{p}}{I_{s}} = \frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$.
$(i)$ If the secondary coil has a greater number of turns than the primary $(N_{s} > N_{p})$,the voltage is stepped up $(V_{s} > V_{p})$. This means the voltage increases in the secondary compared to the primary,and since power is conserved,the current in the secondary becomes less than in the primary $(I_{s} < I_{p})$.
$(ii)$ If the secondary coil has a lesser number of turns than the primary $(N_{s} < N_{p})$,the voltage is stepped down $(V_{s} < V_{p})$. This means the voltage decreases in the secondary compared to the primary,and the current in the secondary becomes more than in the primary $(I_{s} > I_{p})$.

(N/A) $(i)$ Flux Leakage: Not all of the flux due to the primary coil passes through the secondary coil due to poor design of the core or air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other.
$(ii)$ Resistance of the windings: The wire used for the winding has some resistance,and so energy is lost due to heat produced in the wire $(I^{2}R)$. In high current,low voltage windings,these are minimized by using thick wire.
$(iii)$ Eddy Currents: The alternating magnetic flux induces eddy currents in the iron core,causing heating. This effect is reduced by using a laminated core.
$(iv)$ Hysteresis: The magnetization of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material with low hysteresis loss.
$(v)$ Humming/Buzzing: Transformers work on $AC$ voltage; hence,in each cycle,the dimensions of the core change due to magnetostriction. As a result,the transformer produces a buzzing noise,leading to the dissipation of some electrical energy as sound.

(N/A) Transformers are essential for the large-scale transmission and distribution of electric energy over long distances.
The voltage output from the power generator is stepped up using a step-up transformer. This increase in voltage results in a corresponding decrease in current for a given power level $(P = VI)$,which significantly reduces the $I^{2}R$ power loss (heat loss) in the transmission lines.
The high-voltage electricity is then transmitted over long distances to area sub-stations located near the consumers.
At these sub-stations,the voltage is stepped down to safer levels. It is further stepped down at local distribution sub-stations and utility poles before a standard power supply of $240 \ V$ reaches our homes.

(N/A) $(1)$ Small transformers are used in radio receivers,telephones,and loudspeakers.
$(2)$ They are used in TVs,refrigerators,air-conditioners,and computers,and also function as voltage regulators.
$(3)$ They are used for stabilizing power supplies.
$(4)$ Step-up transformers are used to generate high voltage for long-distance power transmission.
$(5)$ Step-down transformers are used for melting metals in induction furnaces.
$(6)$ Step-up transformers are used in the production of $X$-rays.
$(7)$ Electrical energy generated at power stations is transmitted to consumers through transformers.

(N/A) transformer works on the principle of $Mutual \text{ } Induction$.
According to this principle, when the magnetic flux linked with a coil changes, an induced electromotive force $(EMF)$ is produced in the neighboring coil.
In a transformer, an alternating current $(AC)$ flowing through the primary coil creates a time-varying magnetic flux in the core, which is linked to the secondary coil, thereby inducing an alternating $EMF$ in the secondary coil.

(N/A) soft iron core is used in a transformer for two primary reasons:
$1$. High Magnetic Permeability: Soft iron has high magnetic permeability,which allows it to easily magnetize and demagnetize. This ensures that the magnetic flux produced by the primary coil is efficiently linked to the secondary coil,minimizing flux leakage.
$2$. Low Hysteresis Loss: Soft iron has a narrow hysteresis loop,which means the energy dissipated as heat during each cycle of magnetization and demagnetization (hysteresis loss) is very low. This increases the overall efficiency of the transformer.

(N/A) The two types of transformers based on their construction and function are:
$1$. $Step-up$ transformer: This transformer increases the voltage of an alternating current $(AC)$ supply. In this,the number of turns in the secondary coil $(N_s)$ is greater than the number of turns in the primary coil $(N_p)$,i.e.,$N_s > N_p$.
$2$. $Step-down$ transformer: This transformer decreases the voltage of an alternating current $(AC)$ supply. In this,the number of turns in the secondary coil $(N_s)$ is less than the number of turns in the primary coil $(N_p)$,i.e.,$N_s < N_p$.

(B) The transformation ratio,often denoted by $k$ or $r$,is defined as the ratio of the number of turns in the secondary coil $(N_s)$ to the number of turns in the primary coil $(N_p)$.
Mathematically,it is expressed as $k = \frac{N_s}{N_p}$.
For an ideal transformer,this ratio is also equal to the ratio of the secondary voltage $(V_s)$ to the primary voltage $(V_p)$,i.e.,$k = \frac{V_s}{V_p} = \frac{I_p}{I_s}$.

(B) In an ideal transformer, there are no energy losses due to resistance, hysteresis, or eddy currents. According to the principle of conservation of energy, the power input must equal the power output. Therefore, $P_{in} = P_{out}$. The power cannot increase or decrease. Thus, the statement is false.

(B) In an ideal transformer, the power input equals the power output: $P_{in} = P_{out}$.
Since $P = V \times I$, we have $V_p I_p = V_s I_s$.
For a step-down transformer, the secondary voltage is less than the primary voltage $(V_s < V_p)$.
To maintain the equality $V_p I_p = V_s I_s$, the secondary current must be greater than the primary current $(I_s > I_p)$.
Therefore, the output current increases, not decreases. The statement is false.

(0.27 A) Given: Load power $P_{S} = 60\, W$,Secondary current $I_{S} = 0.54\, A$,Primary voltage $V_{P} = 220\, V$ (standard line voltage).
Step $1$: Calculate the secondary voltage $V_{S}$.
$P_{S} = V_{S} I_{S} \implies V_{S} = \frac{P_{S}}{I_{S}} = \frac{60}{0.54} \approx 111.1\, V \approx 110\, V$.
Step $2$: Determine the transformer type.
Since $V_{S} < V_{P}$ $(110\, V < 220\, V)$,the transformer is a step-down transformer.
Step $3$: Calculate the primary current $I_{P}$ using the ideal transformer relation $V_{P} I_{P} = V_{S} I_{S}$.
$I_{P} = \frac{V_{S} I_{S}}{V_{P}} = \frac{60}{220} \approx 0.27\, A$.
Thus,the current in the primary coil is $0.27\, A$ and it is a step-down transformer.

(B) Length of copper wire $L = 2 \times 10\, km = 20000\, m$. Resistance $R = \rho \frac{L}{A} = \frac{1.7 \times 10^{-8} \times 20000}{\pi \times (0.5 \times 10^{-2})^2} \approx 4.33\, \Omega$. Current $I = \frac{P}{V} = \frac{10^6}{220} \approx 4545.45\, A$. Power loss $P_{loss} = I^2 R = (4545.45)^2 \times 4.33 \approx 8.95 \times 10^7\, W$. Since $P_{loss} > P_{transmitted}$,this is not feasible.
$(b)$ New current $I' = \frac{P}{V'} = \frac{10^6}{11000} \approx 90.91\, A$. Power loss $P'_{loss} = (I')^2 R = (90.91)^2 \times 4.33 \approx 35785\, W$. Fraction of power loss $= \frac{35785}{10^6} \approx 0.0358$ or $3.58\%$.

(A) The turns ratio is given by $\frac{N_{1}}{N_{2}} = \frac{50}{1}$.
For an ideal transformer,the voltage ratio is equal to the turns ratio: $\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$.
Given $V_{1} = 120 \ V$,we have $\frac{120}{V_{2}} = \frac{50}{1}$.
Solving for $V_{2}$,we get $V_{2} = \frac{120}{50} = 2.4 \ V$.
The power output is dissipated across the secondary resistance $R_{2} = 1 \ \Omega$.
Using the formula $P_{\text{out}} = \frac{V_{2}^{2}}{R_{2}}$,we substitute the values:
$P_{\text{out}} = \frac{(2.4)^{2}}{1} = 5.76 \ W$.

(B) For an ideal transformer,the power input equals the power output,which implies that the ratio of the number of turns is inversely proportional to the ratio of the currents.
The relationship is given by: $\frac{N_p}{N_s} = \frac{i_s}{i_p}$
Given:
$N_p = 140$
$N_s = 280$
$i_p = 4 \ A$
Substituting the values into the formula:
$\frac{140}{280} = \frac{i_s}{4}$
$\frac{1}{2} = \frac{i_s}{4}$
$i_s = \frac{4}{2} = 2 \ A$
Therefore,the current in the secondary coil is $2 \ A$.

(A) Given:
Number of turns in primary coil,$N_{p} = 500$
Number of turns in secondary coil,$N_{s} = 10$
Load resistance,$R = 10\, \Omega$
Voltage across secondary coil,$V_{s} = 50\, V$
Step $1$: Calculate the current in the secondary coil $(I_{s})$.
Using Ohm's law,$I_{s} = \frac{V_{s}}{R} = \frac{50\, V}{10\, \Omega} = 5\, A$.
Step $2$: Use the transformer ratio to find the current in the primary coil $(I_{p})$.
For an ideal transformer,the ratio of currents is inversely proportional to the ratio of turns:
$\frac{I_{p}}{I_{s}} = \frac{N_{s}}{N_{p}}$
$I_{p} = I_{s} \times \left( \frac{N_{s}}{N_{p}} \right)$
$I_{p} = 5\, A \times \left( \frac{10}{500} \right)$
$I_{p} = 5 \times \frac{1}{50} = \frac{1}{10} = 0.1\, A$.

(C) For an ideal transformer,the input power is equal to the output power.
$P_{\text{in}} = P_{\text{out}}$
$E_{p} I_{p} = \frac{V_{s}^{2}}{R_{s}}$
Substituting the given values:
$1000 \times 50 = \frac{220^{2}}{R_{s}}$
$50000 = \frac{48400}{R_{s}}$
$R_{s} = \frac{48400}{50000} = 0.968 \, \Omega$
Rounding to the nearest integer,we get $R_{s} \approx 1 \, \Omega$.

(C) Given,primary voltage $V_P = 220 \ V$,secondary power $P_S = 60 \ W$,and secondary current $I_S = 0.11 \ A$.
The secondary voltage $V_S$ is given by the formula $V_S = \frac{P_S}{I_S}$.
Substituting the values,$V_S = \frac{60}{0.11} \approx 545.45 \ V$.
Since $V_S > V_P$ $(545.45 \ V > 220 \ V)$,the secondary voltage is greater than the primary voltage.
Therefore,the transformer is a step-up transformer.

(C) The transformer follows the turns ratio formula: $\frac{N_P}{N_S} = \frac{V_P}{V_S}$.
Given:
Primary voltage $V_P = 220\,V$
Secondary voltage $V_S = 12\,V$
Number of turns in secondary $N_S = 24$
Substituting the values into the formula:
$\frac{N_P}{24} = \frac{220}{12}$
$N_P = \frac{220 \times 24}{12}$
$N_P = 220 \times 2 = 440$.
Therefore,the number of turns in the primary coil is $440$.

(A) Given that the transformer is ideal (ignoring power losses),the input power is equal to the output power.
$P_{\text{in}} = P_{\text{out}}$
Since $P_{\text{out}} = 44 \, W$ and $V_p = 220 \, V$,we use the relation $P_{\text{in}} = V_p \times I_p$.
$220 \times I_p = 44$
$I_p = \frac{44}{220} \, A$
$I_p = \frac{1}{5} \, A = 0.2 \, A$
Therefore,the current in the primary circuit is $0.2 \, A$.

(C) For an ideal transformer,the power delivered to the load is the same as the power drawn from the source.
Given: Primary voltage $V_p = 8\,kV = 8000\,V$,Secondary voltage $V_s = 160\,V$,Power $P = 80\,kW = 80000\,W$.
The resistance in the primary circuit $R_p$ is given by $P = \frac{V_p^2}{R_p}$.
$R_p = \frac{V_p^2}{P} = \frac{(8000)^2}{80000} = \frac{64 \times 10^6}{8 \times 10^4} = 8 \times 10^2 = 800\,\Omega$.
The resistance in the secondary circuit $R_s$ is given by $P = \frac{V_s^2}{R_s}$.
$R_s = \frac{V_s^2}{P} = \frac{(160)^2}{80000} = \frac{25600}{80000} = 0.32\,\Omega$.
Thus,the resistances are $800\,\Omega$ and $0.32\,\Omega$.

(C) transformer works on the principle of electromagnetic induction, which requires a time-varying magnetic flux to induce an electromotive force $(EMF)$ in the secondary coil.
The primary coil is connected to a battery, which provides a constant direct current $(DC)$ voltage of $1.5 \, V$.
Since the current is constant, the magnetic flux linked with the primary coil remains constant.
Because the magnetic flux does not change with time, there is no change in the magnetic flux linked with the secondary coil $(\frac{d\Phi}{dt} = 0)$.
According to Faraday's law of induction, the induced $EMF$ in the secondary coil is given by $e = -N \frac{d\Phi}{dt}$. Since $\frac{d\Phi}{dt} = 0$, the induced voltage across the secondary coil is $0 \, V$.

(C) The transformer equation relating the number of turns and voltage is given by: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Given:
Primary voltage $(V_p)$ = $220 \,V$.
Ratio of primary to secondary turns $(\frac{N_p}{N_s})$ = $11:50$,which means $\frac{N_s}{N_p} = \frac{50}{11}$.
Substituting these values into the formula:
$V_s = V_p \times \frac{N_s}{N_p}$
$V_s = 220 \times \frac{50}{11}$
$V_s = 20 \times 50 = 1000 \,V$.
Therefore,the output voltage in the secondary is $1000 \,V$.

(D) Given:
Primary voltage,$V_p = 12\,kV = 12 \times 10^3\,V$
Secondary voltage,$V_s = 120\,V$
Power consumption,$P_s = 60\,kW = 60 \times 10^3\,W$
For an ideal transformer,the power in the secondary circuit is given by $P_s = V_s \times I_s$.
Therefore,the secondary current $I_s = \frac{P_s}{V_s} = \frac{60 \times 10^3}{120} = 500\,A$.
Using Ohm's law for the secondary resistive load $R_s$:
$R_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.24\,\Omega$.
Converting to $m\Omega$:
$R_s = 0.24 \times 10^3\,m\Omega = 240\,m\Omega$.

(B) For an ideal transformer, the input power is equal to the output power.
$P_{in} = P_{out}$
Since $P_{out} = 60\,W$ (power of the lamp) and $P_{in} = V_P \times I_P$, where $V_P = 220\,V$ is the primary voltage.
$220 \times I_P = 60$
$I_P = \frac{60}{220}$
$I_P \approx 0.27\,A$
Therefore, the current in the primary winding is $0.27\,A$.