PHY294H, Assignment 10 - Solutions

FGT 34.8

A V₁ = 220V AC source is applied to the primary of a step down transormer with turn ratio N₁/N₂ = 5. To solve this problem, we assume an ideal transformer. The voltage across the secondary is then five times lower than the voltage on the primary, the current in the secondary is five times higher than the current in the primary. The power dissipated in the secondary is the same as the power delivered by the source which is attached to the primary.

a) The voltage on the secondary is V₂ = N₂ V₁/N₁ = 44V.

b) If a $R_2 = 40\Omega$ resistor is placed on the secondary, the current in the secondary is I₂ = V₂/R₂ = 1.1A. The current in the primary is then I₁ = N₂ I₂/N₁ = 0.22A.

c) The power dissipated in the secondary is P₂ = I₂² R₂. If a resistor R₁ were attached directly to the source, the dissipation would be P₁ = I₁² R₁. If we equate these two powers, we find, $R_1 = I_2^2 R_2/I_1^2 = 1000\Omega$

FGT 34.54

This question is a bit of a lowball. It concerns parallel LRC circuits which we did not cover in class. Nevertheless it illustrates an important point about AC circuits. First we need to find the size of the inductance of the machine shop. We find the resistance from,

$$P = {\cal{E}} I cos(\phi) = 220\times 120 cos(40^o) = 20.2kW$$ (1)

The power is also equal to I²R, so we have,

$$R = P/I^2 = 1.4\Omega$$ (2)

Also note that V=IZ, so $Z=1.83\Omega = (R^2+(X_L-X_C)^2)^{1/2}$.

a) Now we assume that the machine shop is purely inductive, so we have,

$$tan(\phi) = {X_L - X_C\over R} \rightarrow {\omega L \over R}$$ (3)

which implies that $L = R tan(\phi)/\omega = 3.1mH$.

Now we add a capacitor which is in parallel with this inductor. If we choose the capacitor so that the inductor and capacitor are in resonance, then the phase difference between the source voltage and the current which comes out of the source goes to zero. The resonant condition is $\omega = 1/(LC)^{1/2}$, so that the size of the capacitor which we should attach to the circuit is,

$$C = 1/(\omega^2L) = 2.25mF$$ (4)

b) An important feature of the ideal parallel LC combination at resonance is that it has infinite impedance. This is completely the opposite of the series LC combintation which has zero impedance at resonance. This difference is due to the fact that the (ideal) inductor and capacitor have voltages which want to be out of phase. They can achieve this in a series combintation, but in a parallel combination, Kirchhoff's loop law states that they must be in phase. The fact that the source must force them to be in phase in the parallel combination leads to a large impedance. Since the impedance of the ideal parallel LC combintation diverges at resonance, we have

$$I (at resonance) = 220/Z \rightarrow 0$$ (5)

That is, the current in the main cable is very small, and this reduces the losses in the cable.

c) The power used by the machine shop is very small (for ideal inductor and capacitor). In real life the machines themselves have a resistance as does the capacitor. This actually reduces the effective impedance of the LC combination.

FGT 34.68

The DC component of the source is blocked by the capacitor. The AC component has an impedance which is determined by the frequency R and C. We consider the AC component only and assign it amplitude V_in.

The ratio of the voltage across the resistor ( V_out = I R), with voltage supplied by the source (V_in=IZ) is given by,

$${V_{out} \over V_{in}} = {R\over Z} = {R\over [R^2+({1\over \omega C})^2]^{1/2}} = {\omega RC \over [1+(\omega R C)^2]^{1/2}}$$ (6)

At low frequencies this ratio approaches zero, while at high frequencies it approaches one. It is a high pass filter as for high frequencies all of the applied voltage is passed to the resistor. It blocks the voltage at low frequencies as there the capacitor has high impedance and so most of the voltage is dropped across the capacitor. At high frequencies the capacitor is like a short circuit, so all of the voltage is dropped across the resistor.

FGT 34.82

The voltage across the capacitor is given by V_C = IX_C. If the maximum voltage that the capacitor can withstand is 1200V (peak), then the maximum (peak) current which the source should supply is,

$$I = 1200/X_C = 1200\omega C = 15.8A$$ (7)

The impedance of the circuit at this current is given by,

$$Z = {V\over I} = {2^{1/2} 220\over 15.8} = 19.65\Omega$$ (8)

The impedance is given by Z² = R² + (X_L-X_C)², so we have,

$$X_L = \omega L =\pm(Z^2 - R^2)^{1/2} + X_C = (92.7\Omega,58.87\Omega)$$ (9)

Which implies that $L=X_L/\omega = (0.245H,0.156H)$. Provided the inductance does not lie between these two values, the voltage across the capacitor remains below 1200V. Another way of thinking about this problem is to plot the current as a function of inductance. The resonant condition gives $L_{res} = 1/(\omega^2C) = 0.201H$. Near resonance the impedance is small, the current is large (I=V/Z), so the voltage across the capacitor (V_C=IX_C) is large. The range of inductances between 156mH and 245mH is close to resonance for this circuit and so in this range of inductances the voltage across the capacitor is large. Outside this range of inductances, the capacitor survives.