Your Answer : (Not Answered)

Correct Answer : D

The mho relay is usually employed for long transmission lines.

The mho (admittance) relay is a distance relay whose operating characteristic is a circle through the origin on the impedance diagram. It is inherently directional and covers the smallest area of the R-X plane for a given reach, so it is the least affected by power swings, which are a serious problem on long, heavily loaded lines.

The usual textbook matching is: reactance relay for short lines, where arc resistance would upset other relays; impedance relay for medium lines; mho relay for long lines. Transformers are protected mainly by differential relays, not distance relays.

Your Answer : (Not Answered)

Correct Answer : B

It is called the recovery voltage.

When a circuit breaker interrupts a fault, the arc is extinguished at a current zero, and the system voltage then reappears across the open contacts. The power-frequency voltage that appears across the contacts after final arc extinction is the recovery voltage; the high-frequency transient that rides on it immediately after interruption is the restriking voltage.

Arc voltage is the drop across the arc while it is still burning, before interruption. A surge voltage is a transient from lightning or switching, not the steady voltage across open contacts, and 'break open voltage' is not a standard term.

Your Answer : (Not Answered)

Correct Answer : D

The correct relation is \(\frac{n_{0}}{N} = \frac{(I_{1} + I_{2})/2}{I_{1} - I_{2}}\).

In a biased (percentage) differential relay, the operating coil carries the difference current \(I_{1} - I_{2}\), and the restraining coil, connected at the midpoint of the pilot wires, carries the average current \(\frac{I_{1} + I_{2}}{2}\). At the operating threshold the two sets of ampere-turns balance:

\[ \begin{aligned} n_{0}(I_{1} - I_{2}) &= N\,\frac{I_{1} + I_{2}}{2} \\ \frac{n_{0}}{N} &= \frac{(I_{1} + I_{2})/2}{I_{1} - I_{2}} \end{aligned} \]

The first option is this ratio turned upside down. The two options with \(n_{0} \times N\) multiply the turns, which does not come from balancing ampere-turns.

Your Answer : (Not Answered)

Correct Answer : D

The rated breaking capacity equals the product of rated symmetrical breaking current and rated voltage.

The breaking capacity of a circuit breaker is stated in MVA and is based on the rms value of the symmetrical (ac) component of the current it can interrupt at its rated voltage. For a three phase breaker the product carries the factor √3: breaking capacity = √3 × rated voltage × rated symmetrical breaking current.

The other options are less precise or wrong. 'Breaking current' without 'symmetrical' leaves out which component is meant, and the fault voltage is not the basis of the rating. 'Twice the rated current and voltage' describes no standard rating.

Your Answer : (Not Answered)

Correct Answer : C

The answer is Statement I is false but Statement II is true.

Statement I is false: an air blast breaker's extinguishing power is the same whatever the current. When it interrupts a small current, such as the magnetising current of an unloaded transformer, the arc is cut off before current zero (current chopping), causing high overvoltages. It is suited to interrupting large currents at high voltage quickly, not to low-magnitude currents.

Statement II is true: minimum oil breakers, which use a small quantity of oil only for arc quenching, are used for medium and high voltages up to about the 220 kV level; above that, air blast and SF₆ breakers take over.

Your Answer : (Not Answered)

Correct Answer : B

The asymmetrical breaking current is \(\sqrt{I_{dc}^{2} + \left(\frac{I_{m}}{\sqrt{2}}\right)^{2}}\).

The asymmetrical breaking current is the rms value of the total current at the instant the contacts separate. That current is the sum of a dc component \(I_{dc}\) and an ac component whose rms value is \(I_{m}/\sqrt{2}\). The rms value of a dc plus a sinusoid adds in squares:

\[ \begin{aligned} I_{ac} &= \frac{I_{m}}{\sqrt{2}} \\ I_{asymm} &= \sqrt{I_{dc}^{2} + I_{ac}^{2}} \end{aligned} \]

Adding the two components directly overstates the rms value. Leaving out the square root gives a quantity in amperes squared, and taking the root of the plain sum has the wrong units as well.

Your Answer : (Not Answered)

Correct Answer : A

The primary function of a circuit breaker is to protect the circuit from overloading and faults.

A circuit breaker is a switch that can make and break the circuit under normal conditions and, crucially, interrupt the large currents of an overload or short circuit, isolating the faulty section before the equipment is damaged.

It does not energise the relay; the relay senses the fault and energises the breaker's trip coil. Measuring fault current is done by current transformers and relays, and raising voltage is the job of a transformer.

Your Answer : (Not Answered)

Correct Answer : A

Relay co-ordination refers to the timing of protection relays.

In a network, several relays see the same fault current. Co-ordination sets their current and time settings so that the relay nearest the fault trips first and isolates the smallest part of the system, while relays further back wait a grading interval, typically 0.3 to 0.5 s, and act only as back-up if it fails.

It has nothing to do with balancing load or placing current transformers. Mal-operation of relays is what good co-ordination prevents, not what the term means.

Your Answer : (Not Answered)

Correct Answer : D

A mho relay is a distance relay with a circular characteristic that passes through the origin, and it compares an operating quantity with a restraining quantity, with a polarising input that gives it direction. So its inputs are the operating input, the restraining input and the polarising input. Galvanic input is not one of the inputs of a mho relay (galvanic isolation is a property of circuits and not a relay input). The answer is option D.

Your Answer : (Not Answered)

Correct Answer : B

An impedance relay is an over current relay with a voltage restraint. In its operation the current produces the operating (positive) torque, tending to close the contacts, while the voltage produces the restraining torque, which opposes the closing. So the voltage produces a negative torque, and the relay operates when the ratio of voltage to current, the impedance, falls below the setting. The answer is option B.

Your Answer : (Not Answered)

Correct Answer : A

The two sources drive a current through the series path \(Z_A + Z_L + Z_B = 150\ \Omega\). With \(E_A\) and \(E_B\) in phase,

\[ I = \frac{430 - 330}{150} = 0.667\ \text{A} \]

The relay voltage is that of the bus after \(Z_A\):

\[ \begin{aligned} V = E_A - IZ_A = 430 - 0.667 \times 50 \\ &= 396.7\ \text{V} \end{aligned} \]

\[ \begin{aligned} Z_{relay} = \frac{V}{I} = \frac{396.7}{0.667} \\ &= 595\ \Omega \end{aligned} \]

Equivalently, \(Z = Z_L + Z_B + E_B/I = 100 + 495 = 595\ \Omega\). The answer is option A.

Your Answer : (Not Answered)

Correct Answer : C

With differential protection, only 85 per cent of the winding is protected against a phase to ground fault, and the relay operates when the out of balance current is 20 per cent of the full load current.

\[ \begin{aligned} I_{FL} \\ &= \frac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} \\ &= 5249\ \text{A} \end{aligned} \]

\[ \begin{aligned} I_{relay} = 0.2 \times 5249 \\ &= 1049.8\ \text{A} \approx 1.049\ \text{kA} \end{aligned} \]

The primary earth fault current at which the relay operates is therefore about 1.049 kA. The values of 10.49 kA, 0.1049 kA and 104.9 kA are slips of a power of ten. The answer is option C.

Your Answer : (Not Answered)

Correct Answer : A

The key gives 0.128 s. For an extremely inverse over current relay the operating time is inversely proportional to the square of the plug setting multiplier, \(t = \dfrac{80\,\text{TMS}}{PSM^2 - 1}\) in the standard IEC form. With the fault current of 25 A and the time multiplier setting of 100 per cent, the plug setting multiplier depends on a relay setting that the item does not print, so the stated data do not let the value be derived independently. The explanation follows the key, option A, and the answer is of the order of a fraction of a second, which is typical for an extremely inverse curve at a high multiple of the setting.

Your Answer : (Not Answered)

Correct Answer : D

The common protective schemes in a power system are over current protection, distance (impedance) protection, differential protection, carrier-current (pilot) protection and earth fault protection. Each responds to an electrical quantity such as current, impedance or a signal sent over the line. Humidity is a condition of the air and not an electrical quantity that a relay measures, so humidity protection is not a type of protective scheme. The answer is option D.

Your Answer : (Not Answered)

Correct Answer : C

The plug setting of an over current relay is a percentage of the rated secondary current of the current transformer. A CT with a ratio of 400/8 has a rated secondary current of 8 A.

\[ \begin{aligned} \text{Pick up current} \\ &= 12.5\% \times 8\ \text{A} = 0.125 \times 8 \\ &= 1\ \text{A} \end{aligned} \]

This is the current in the relay coil at which it picks up; the primary current that causes it is \(1 \times 50 = 50\) A, because the CT ratio is 50. The values 5 A, 2.5 A and 0.5 A do not match 12.5% of 8 A. The answer is 1 A, option C.

Your Answer : (Not Answered)

Correct Answer : D

The advantages of an HRC (high rupturing capacity) fuse are that it needs no maintenance, it operates very fast and so limits the let through current of a short circuit, and it is cheap. Its disadvantage is that it is a one shot device: once the element has blown, the whole fuse link must be replaced, so it cannot be reused for multiple operations, and the circuit breaker, in contrast, can be reclosed. So the reuse statement is the one that is not an advantage. The answer is option D.

Your Answer : (Not Answered)

Correct Answer : B

An expulsion type lightning arrester has a spark gap in series with a fibre tube that produces gas when the arc passes and so blows the arc out. Its breaking capacity and the limits of its protection make it a device for distribution systems, and it is commonly used on systems up to 33 kV. For higher voltages, such as 66 kV or 132 kV, valve type or metal oxide arresters are used, and 3.3 kV is too low a limit for it. The answer is 33 kV, option B.