EE MCQ

SSC JE Electrical 2018 with solution SET-2
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What will be the value of Thevenin’s voltage (in V). The Thevenin’s resistance (in Ω) and the load current (in A) respectively, across the load resistor in the given electrical circuit?

As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage ETh and internal resistance RT and this is called Thevenin equivalent circuit.

Thevenin Voltage

The Thevenin voltage used in Thevenin’s Theorem is an ideal voltage source equal to the open circuit voltage at the terminals.

In the given question, the resistance 10Ω does not affect this voltage and the resistances 30Ω and 20Ω form a voltage divider, giving

\begin{array}{l}{E_{Th}} = 100 \times \dfrac{30}{{30 + 20}}\\\\{E_{TH}} = 60V\end{array}

Thevenin’s resistance can be found by replacing 100 V source with a short-circuit.

Thevenin equivalent resistance for the given network is

R = (R1 || R2) + R3

Rth = (20 || 30) + 10 = (20 x 30) ⁄ (20 + 30) + 10 = 22Ω

Rth = 22Ω

The Load current Is calculated as

IL = ETH ⁄ (RTH + RL)

= 60 ⁄ (22 + 5) = 2.22 A

Hence the value of Thevenin voltage, Thevenin Resistance, and Load current is (60 V, 22Ω, 2.22A) respectively.

Related Question

SSC JE Electrical 2018 with solution SET-2
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Determine the value of current (in A) through both the resistor of the given circuit

Current through the 10Ω resistance

I1 = V/R = 20/10 = 2A

I1 = 2A

Now current through the 20Ω resistance

I2 = V − (-10)/R = 20 + 10/30 = 1.5 A

I2 = 1.5 A

SSC JE Electrical 2018 with solution SET-2
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Which of the following law is based on the conservation of energy?

Kirchhoff’s Voltage Law (KVL,) or Kirchhoff’s Loop Rule. This law is based on the conservation of energy and may be stated as under:

In any closed electrical circuit or loop, the algebraic sum of all the electromotive force (e.m.f s) and voltage drops in resistors is equal to zero, i.e., in any closed circuit or loop.

The algebraic sum of e.m.f s + Algebraic sum of voltage drops = 0

The validity of Kirchhoff’s voltage law can be easily established by referring to the loop ABCDA shown in Fig.

If we start from any point (say point A) in this closed circuit and go back to this point (i.e., point A) after going around the circuit, then there is no increase or decrease in potential. This means that algebraic sum of the e.m.f.s of all the sources (here only one e.m.f. source is considered) met on the way plus the algebraic sum of the voltage drops in the resistances must be zero. Kirchhoff’s voltage law is based on the law of conservation of energy, i.e., the net change in the energy of a charge alter completing the closed path is zero.

V1 + V2 − V = 0

or

Kirchhoff’s voltage law is also called as loop rule.

SSC JE Electrical 2018 with solution SET-2
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Which of the following statement is CORRECT?

Although the Thevenin’s theorem and Norton’s theorem can be used to solve a given network, yet the circuit approach differs in the following respects:

A Norton’s theorem is converse (opposite) of Thevenin’s theorem in the respect that Norton equivalent circuit uses a current generator instead of the voltage generator and the resistance RN (which is the same as RTH) in parallel with the generator instead of being in series with it.

Thevenin’s theorem is a voltage form of an equivalent circuit whereas Norton’s theorem is a current form of an equivalent circuit.

To Convert Thevenin equivalent circuit into Norton’s equivalent circuit the following step is involved

RN = RTH

IN = ETH ⁄ RTH

SSC JE Electrical 2018 with solution SET-2
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Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.

Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).

RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5

RN = 7.66

The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.

According to voltage Division Rule

VAB = V1R3 ⁄ (R1 + R3)

= 24 × 8 ⁄ (4 + 8)

VAB = 16 V

Now Norton Current IN is

IN = VAB ⁄ RN

IN = 16 ⁄ 7.66

IN = 2.08

SSC JE Electrical 2018 with solution SET-2
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How much power (in W) will be dissipated by a 5 Ohm resistor in which the value of current is 2 A?

Given

Resistance R = 5Ω

Current I = 2 A

Power dissipated by the resistor is

P = I2R

P = 22 × 5

P = 20 watts