Question

(I) An ac source of voltage V=$V_0$ sin $\omega t$ is connected to a series combination of L,C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(ii) In a series LR circuit $X_L=R$ and power factor of the circuit is $P_1$. When capacitor with capacitance C such that $X_L=X_C$ is put in series, the power factor becomes $P_2$. Calculate $P_1$/$P_2$.
OR
(i) Write the function of transformer. state its principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100.The input voltage and power are respectively 220 V and 1100 W. Calculate
(a) number of turns in secondary
(b) voltage across secondary
(c) current in primary
(d) current in secondary
(e) power in secondary

Answer 1

Refer IMAGE $01$ and $02$

$\left(i\right)$ $V=\sqrt{i^2(X_L-X_C{)}^2+i^2{R}^2}$

$i=\frac{V}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

Refer IMAGE $03$

$\left(ii\right)$ When, $X_L=R$

Power Factor $P_1=\cos \varphi$

$\cos \varphi =\frac{IR}{\sqrt{(iR{)}^2+(i{X}_C{)}^2}}=\frac{1}{\sqrt{2}}$

CASE $2$ $X_L-X_C=0$

$P_2=\cos {\varphi }_2=1$

Ratio, $\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}$

OR

$\left(i\right)$ Transformer is used to convert low voltage to high high voltage and high voltage to low voltage as well as high current to low current and low current to high current, at constant power and frequency.

It working on principal of mutual inductance various losses are given below.

$1.$ Eddy current losses

$2.$ Hysteresis Losses (Constant losses)

Refer IMAGE $04$

$\left(ii\right)$ Given that $N_1=100$

Transformation ratio $\frac{N_2}{N_1}=100$

$\left(a\right)$ $N_2=100N_1=10000$ Turns

$\left(b\right)$ $\frac{V_1}{V_2}=\frac{N_1}{N_2}$

$V_2=V_1\frac{N_2}{N_1}=220\times 100=22$ $kV$

$\left(c\right)$ Current Passing in $I$

Power, $P=VI,P_1=V_1{P}_1$

$\Rightarrow 1100=220I$

$\Rightarrow I=\frac{1100}{220}=5$ $A$

$\left(d\right)$ $\frac{I_1}{I_2}=\frac{N_2}{N_1}$

$I_2=\frac{5}{N_2/N_1}=\frac{5}{100}=50$ $mA$

$\left(e\right)$ Power in secondary is constant

$\Rightarrow P_1=P_2=1100$ $W$