EE MCQ

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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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.

SSC JE Electrical 2018 with solution SET-2
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Determine the value of current I1 (in A) and V1 (in V) respectively, for the given circuit below.

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

∴ I1 = 1 + 3 = 4A

V = IR

∴ V1 = I1R = 8 × 4

V1 = 32Ω

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