Question
State the principle on which transformer works. Explain its working with construction. Derive an expression for ratio of e.m.f.s and currents in terms of number of turns in primary and secondary coil.
✓
Answer
A transformer is a device with the help of which, a given alternating voltage can be increased or decreased to any desired value. The first type of transformer which delivers an output voltage smaller than the input voltage
is called a step down transformer. The second type of transformer which delivers an output voltage larger than the input voltage is called a step up transformer.
Principle of working of a transformer:
A transformer works on the principle that whenever the magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.Construction:
It consists of two coils, primary (P) and secondary (S), insulated from each other and wound on a soft iron core as shown in the figure below.
The primary coil is called the input coil and the secondary coil is called the output coil.
Working:
When an alternating voltage is applied to the primary coil, the current through the coil goes on changing. Hence, the magnetic flux through the core also changes. As this changing magnetic flux is linked with both coils, an emf is induced in each of them. The amount of magnetic flux linked with the coil depends on the number of turns of the coil.Derivation:
Let ‘$Φ$’ be the magnetic flux linked per turn with both coils at a certain instant of time $‘t’.$
Let the number of turns of the primary and secondary coils be $‘N_p’$ and $‘N_s’$, respectively.
Therefore, the total magnetic flux linked with the primary coil at certain instant of time $‘t’$ is $N_pΦ$. Similarly, the total magnetic flux linked with the secondary coil at certain instant of time $‘t’$ is $N_sΦ$.
Now, the induced emf in a coil is
\(e=\frac{d \phi}{d t}\)
Therefore, the induced emf in the primary coil is
\(e_p=\frac{d \phi_p}{d t}=\frac{d N_p \phi}{d t}=-N_p \frac{d \phi}{d t}\).......(1)
Similarly, the induced emf in the secondary coil is
\(e_s=\frac{d \phi_s}{d t}=\frac{d N_s \phi}{d t}=-N \frac{d \phi}{d t}\).......(2)
Dividing equations (1) and (2), we get
\(\frac{e_s}{e_p}=\frac{-N_s \frac{d \phi}{d t}}{-N_p \frac{d \phi}{d t}}=\frac{N_s}{N_p}\)........(3)
The above equation is called the equation of the transformer and the ratio \(\frac{N_s}{N_p}\) is known as the turns ratio of the transformer.
Now, for an ideal transformer, we know that the input power is equal to the output power.
\(\therefore P_p=P_s\)
\(\therefore e_p i_p=e_s i_s\)
\(\therefore \frac{e_s}{e_p}=\frac{i_p}{i_s}\)
From equation (3), we have
\(\frac{e_s}{e_p}=\frac{N_s}{N_P}\)
\(\therefore \frac{e_s}{e_p}=\frac{N_s}{N_p}=\frac{i_p}{i_s}\)
is called a step down transformer. The second type of transformer which delivers an output voltage larger than the input voltage is called a step up transformer.
Principle of working of a transformer:
A transformer works on the principle that whenever the magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.Construction:
It consists of two coils, primary (P) and secondary (S), insulated from each other and wound on a soft iron core as shown in the figure below.
The primary coil is called the input coil and the secondary coil is called the output coil.
Working:
When an alternating voltage is applied to the primary coil, the current through the coil goes on changing. Hence, the magnetic flux through the core also changes. As this changing magnetic flux is linked with both coils, an emf is induced in each of them. The amount of magnetic flux linked with the coil depends on the number of turns of the coil.Derivation:
Let ‘$Φ$’ be the magnetic flux linked per turn with both coils at a certain instant of time $‘t’.$
Let the number of turns of the primary and secondary coils be $‘N_p’$ and $‘N_s’$, respectively.
Therefore, the total magnetic flux linked with the primary coil at certain instant of time $‘t’$ is $N_pΦ$. Similarly, the total magnetic flux linked with the secondary coil at certain instant of time $‘t’$ is $N_sΦ$.
Now, the induced emf in a coil is
\(e=\frac{d \phi}{d t}\)
Therefore, the induced emf in the primary coil is
\(e_p=\frac{d \phi_p}{d t}=\frac{d N_p \phi}{d t}=-N_p \frac{d \phi}{d t}\).......(1)
Similarly, the induced emf in the secondary coil is
\(e_s=\frac{d \phi_s}{d t}=\frac{d N_s \phi}{d t}=-N \frac{d \phi}{d t}\).......(2)
Dividing equations (1) and (2), we get
\(\frac{e_s}{e_p}=\frac{-N_s \frac{d \phi}{d t}}{-N_p \frac{d \phi}{d t}}=\frac{N_s}{N_p}\)........(3)
The above equation is called the equation of the transformer and the ratio \(\frac{N_s}{N_p}\) is known as the turns ratio of the transformer.
Now, for an ideal transformer, we know that the input power is equal to the output power.
\(\therefore P_p=P_s\)
\(\therefore e_p i_p=e_s i_s\)
\(\therefore \frac{e_s}{e_p}=\frac{i_p}{i_s}\)
From equation (3), we have
\(\frac{e_s}{e_p}=\frac{N_s}{N_P}\)
\(\therefore \frac{e_s}{e_p}=\frac{N_s}{N_p}=\frac{i_p}{i_s}\)
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