Table of Contents

### Q.1. What is the practical importance of parallel a.c. circuits?

Ans. Parallel a.c. circuits are used more frequently in electrical systems than in other types of a.c. circuits. For example, electrical devices and equipment are connected in parallel across a.c. mains. there are two principal reasons for it.

(i) The operation of each device becomes independent of the other. therefore, it is possible to turn ON or OFF any device without disturbing the operation of other devices.

(ii) Most electrical appliances requiring different currents at the same voltage are to be connected to the same power source. this necessitates parallel connections.

### Q. 2. Why Jo we prefer the use of phasor algebra in solving parallel a.c. circuit problems?

Ans. It is a usual practice to solve parallel a.c. circuit problems by the use of phasor algebra due to the following reasons:

(i) Voltages, currents, and impedances can be expressed in complex notation. This makes the entire treatment very simple.

(ii) In the case of complex circuits, this method yields quick results.

(iii) The need of drawing a phasor diagram is eliminated.

Q. 3. Why do we discourage the use of the equivalent impedance method in solving parallel circuit problems?

Ans. In order to find the equivalent impedance of a parallel circuit in complex form, a lot of lengthy calculations are involved. Such an approach to solving parallel a.c. circuits are recommended particularly when there are more than two branches in the circuit.

Q. 4. What is the advantage of the admittance method in solving parallel a.c. circuit problem?

Ans. The total admittance (in complex form) of a parallel circuit is the sum of individual admittances (in complex form) of the parallel branches. In other words, the admittance of the parallel branches is added. Thus admittance method in a parallel circuit makes the approach somewhat similar to a series circuit where impedances (in complex form) are added.

### Q.5. What is the origin of the name resonance?

Ans. An electrical circuit containing reactive elements (L and C) is said to be in resonance if the circuit power factor is unity. if it happens in a series a.c. circuit (i.e.R-L-C series circuits), it is called series resonance. if this condition (i.e., p.f.) occurs in parallel a.c circuits, it is termed parallel resonance. unity p.f. means that circuit voltage and current are in phase i.e., in step with each other. resonance also means to be in step with.

### Q. 6. If the resistor is shunted across a parallel LC circuit, what is its effect on (i) circuit impedance (ii) circuit (iii) bandwidth, and (iv) gain?

Ans. In practical applications, it sometimes happens that a resistive loading. the effect is shunted across a parallel-resonant circuit. At other times, an actual resistor may be deliberately connected across the parallel-resonant circuit to obtain some desired effects. The effects of shunting a resistive load across a parallel-resonant circuit are :

(1) The impedance of the parallel circuit is decreased.

(ii) The Q of the circuit is decreased.

(iii) The bandwidth of the circuit is increased.

(iv) Since the Q is decreased, the gain is reduced.

### Q. 7. Under what condition will the impedance of a parallel LC circuit appear as a low capacitive reactance?

Ans. When the supply frequency is sufficiently greater than the resonant frequency of a parallel tuned circuit, the circuit behaves as a low capacitive circuit.

### Q. 8. In a parallel LC resonant circuit, I_{L} = I_{C} why is not the line current zero?

Ans. At parallel resonance, the two branch currents I_{L} and l_{C} are equal11 in magnitude. However, since the inductive branch current, I_{L} is not quite at 90^{0} lagging, the phasor sum of branch currents is not zero but it is a minimum value. This line current is in phase with the line voltage. The small Current flowing in the line is only the amount needed to supply the resistance losses.

### Q. 9. What is the power dissipated in admittance?

Ans. Power in admittance, P= VI cos Φ

=V^{2}Y cos Φ (since I = VY)

= V^{2}G (since Y cos Φ = G)

It can be seen that the conductance of the circuit is responsible for power dissipation.