Question

a) State the principle of a transformer. Explain its construction and working. Derive an expression for the ratio of e.m.f.s in terms of number of turns in primary and secondary coil.
b)Two diametrically opposite points of a metal ring are connected to two terminals of the left gap of meter bridge. The resistance of $11\Omega$ is connected in right gap. If null point is obtained at a distance of $45cm$ from the left end, find the resistance of metal ring.

Answer 1

Transformer : 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 : An e.m.f. is induced in the secondary coil due to the changes of current in a neighboring coil.

Construction: A simple transformer is as shown in figure. It consists of a closed, laminated soft iron core on which two coils having different number of turns are wound. The coils are made of insulated wire. The coil to which the A.C. input voltage is applied is called the primary coil and the coil across which the A.C. output voltage is obtained is called the secondary coil.

Working: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. Let $\varphi$ be the flux in each turn in the core at time t due to current in the primary when a voltage $V_p$ is applied to it.

Then the induced emf or voltage ${ϵ}_s$, in the secondary with $N_s$ turns is

${ϵ}_s=-N_s\frac{d\varphi }{dt}$

The alternating flux $\varphi$ also induces an emf, called back emf in the primary. This is ${ϵ}_p=-N_p\frac{d\varphi }{dt}$

But ${ϵ}_p=V_p$ If this were not so, the primary current would be infinite since the primary has zero resistance(as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation ${ϵ}_s=V_s$ where $V_s$ is the voltage across the secondary.

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

b) Let $R$ be the resistance of metal ring.

Equivalent resistance in left gap is given by $\frac{1}{R_p}=\frac{1}{R/2}+\frac{1}{R/2}$

$R_p=\frac{R}{4}$

So, $\frac{R_p}{11}=\frac{45}{55}\Rightarrow R=\frac{45\times 11\times 4}{55}$
$\Rightarrow R=36\Omega$