A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of tansformer being used.
Given, $P_S=60 \mathrm{~W}, I_S=0.54 \mathrm{~A}$
Current in the primary $I_p=$ ?
Taking line voltage as 220 V.
We can write Since,
$$\begin{array}{ll} \Rightarrow & P_L=60 \mathrm{~W}, I_L=0.54 \mathrm{~A} \\ \Rightarrow & V_L=\frac{60}{0.54}=110 \mathrm{~V} .\quad\text{.... (i)} \end{array}$$
Voltage in the secondary $\left(E_S\right)$ is less than voltage in the primary $\left(E_P\right)$.
Hence, the transformer is step down transformer.
Since, the transformation ratio
$r=\frac{V_s}{V_p}=\frac{I_p}{I_s}$
$$\begin{array}{lr} \text { Substituting the values, } & \frac{110 \mathrm{~V}}{220 \mathrm{~V}}=\frac{I_p}{0.54 \mathrm{~A}} \\ \text { On solving } & I_p=0.27 \mathrm{~A} \end{array}$$
Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge).
Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency. Mathematically, the reactance can be written as $X_C=\frac{1}{\omega C}$.
Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
An inductor opposes flow of current through it by developing a back emf according to Lenz's law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice -versa.
Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency. Mathematically, the reactance offered by the inductor is given by $X_L=\omega L$.
7 An electrical device draws 2 kW power from AC mains (voltage 223 V $(\mathrm{rms})=\sqrt{50000} \mathrm{~V}$ ). The current differs (lags) in phase by $\phi\left(\tan \phi=\frac{-3}{4}\right)$ as compared to voltage. Find (a) $R$, (b) $X_C-X_L$ and (c) $I_M$. Another device has twice the values for $R, X_C$ and $X_L$. How are the answers affected?
$$\begin{aligned} & \text { Given, power drawn }=P=2 \mathrm{~kW}=2000 \mathrm{~W} \\ & \tan \phi=-\frac{3}{4}, I_M=I_0=\text { ?, } R=\text { ?, } X_C-X_L=\text { ? } \\ & V_{\mathrm{rms}}=V=223 \mathrm{~V} \\ & \text { Power } P=\frac{V^2}{Z} \\ \Rightarrow\quad& Z=\frac{V^2}{P}=\frac{223 \times 223}{2 \times 10^3}=25 \end{aligned}$$
$$\begin{aligned} \text { Impedance } Z & =25 \Omega \\ \text { Impedance } Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ \Rightarrow\quad 25 & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ \text{or}\quad 625 & =R^2+\left(X_L-X_C\right)^2 \quad\text{... (i)}\\ \text{Again}\quad\tan \phi & =\frac{X_L-X_C}{R}=\frac{3}{4} \\ \text{or}\quad X_L-X_C & =\frac{3 R}{4}\quad\text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { From Eq. (ii), we put } X_L-X_C=\frac{3 R}{4} \text { in Eq. (i), we get }\\ &625=R^2+\left(\frac{3 R}{4}\right)^2=R^2+\frac{9 R^2}{16} \end{aligned}$$
or $625=\frac{25 R^2}{16}$
(a) Resistance $R=\sqrt{25 \times 16}=\sqrt{400}=20 \Omega$
(b) $X_L-X_C=\frac{3 R}{4}=\frac{3}{4} \times 20=15 \Omega$
(c) Main current $I_M=\sqrt{2} I=\sqrt{2} \frac{V}{Z}=\frac{223}{25} \times \sqrt{2}=12.6 \mathrm{~A}$
As $R, X_C, X_L$ are all doubled, $\tan \phi$ does not change. $Z$ is doubled, current is halved. So, power is also halved.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if
(i) power is transmitted at 220 V . Comment on the feasibility of doing this.
(ii) a step-up transformer is used to boost the voltage to 11000 V , power transmitted, then a step-down transformer is used to bring voltage to 220 V.
$$\left(\rho_{\mathrm{cu}}=1.7 \times 10^{-8} \text { SI unit }\right)$$
(i) The town is 10 km away, length of pair of Cu wires used, $L=20 \mathrm{~km}=20000 \mathrm{~m}$.
Resistance of Cu wires,
$$\begin{aligned} R & =\frac{l}{A}=\frac{l}{\pi(r)^2} \\ & =\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^2}=4 \Omega \end{aligned}$$
$$\begin{aligned} I \text { at } 220 \mathrm{~V} \quad V I & =10^6 \mathrm{~W} ; I=\frac{10^6}{220}=0.45 \times 10^4 \mathrm{~A} \\ R I^2 & =\text { power loss } \\ & =4 \times(0.45)^2 \times 10^8 \mathrm{~W} \\ & >10^6 \mathrm{~W} \end{aligned}$$
Therefore, this method cannot be used for transmission.
(ii) When power $P=10^6 \mathrm{~W}$ is transmitted at 11000 V .
$$\begin{aligned} V^{\prime} I^{\prime} & =10^6 \mathrm{~W}=11000 I^{\prime} \\ \text { Current drawn, } I^{\prime} & =\frac{1}{1.1} \times 10^2 \\ \text { Power loss } & =R I^2=\frac{1}{1.21} \times 4 \times 10^4 \\ & =3.3 \times 10^4 \mathrm{~W} \\ \text { Fraction of power loss } & =\frac{3.3 \times 10^4}{10^6}=3.3 \% \end{aligned}$$