# (a)  Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.(b)  A small town with a demand of  $1200\; kW$ of electric power at  $220\; V$ is situated $\inline 20\; km$ away from an electric plant generating power at $440\; V$. The resistance of the two wire line carrying power is $\inline 0.5\; \Omega\; per\; km$. The town gets the power from the line through a $\inline 4000-220\; V$.  step- down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.

(a)    A device which is used to decrease high ac voltage is known as a step- down transformer,

It works on the principle of mutual induction, It states that, if the current is changing through the primary coil, then there is a change in magnetic flux in the secondary coil. This change in flux induces an emf in the secondary coil.

The four sources of energy loss in the device are :

Flux leakage, eddy currents, Resistance of the windings and the hysteresis loss.

(b) Given,

The  town has a power demand of $\inline 12000\; kw$ at $\inline 220 \; V$

Resistance per unit length $\inline =0.5\; \Omega /km$

Length of each wire $\inline =20 \; km$

Total resistance, $\inline r=2\times 0.5\times 20\; \Omega =20\; \Omega$

The transformers step down from $4000 V (V_1)$  to  $\inline 220\; V(V_{2})$ .

If there is no power loss in the transformers, then we can calculate the RMS current in the lines by :

$I=\frac{P}{V_{1}}=\frac{1200\times 10^{3}w}{4000v}=300A$

The power loss in the lines is

$P_{loss}=I^{2}r$

$=(300\; A)^{2}\times 20\; \Omega$

$=1800000\; w$

$=1800\; kw$

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