EE MCQ

SSC JE Electrical 2019 with solution SET-1
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The Norton’s current in the circuit shown below is

According to the Norton theorem, to find the Norton current, first remove the load resistance RL from the network terminals AB. Short circuit the terminals AB as shown in Figure calculate the current ISc or IN through the short circuit.

Now resistance of 150Ω will not show any effect in the circuit. So only resistance of 30Ω will be effective.

Norton current IN = 360/30

IN = 12 A

Related Question

SSC JE Electrical 2019 with solution SET-1
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The direction of the arrow represents the direction of when the diode is forward biased

The direction of the arrow represents the direction of the conventional current flow when the diode is forward biased.

SSC JE Electrical 2019 with solution SET-1
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The output of the logic circuit given below represents the gate.

In the given diagram all are NOR Gate . The final output is shown in the figure.

At stage 1 the output will be \overline A \& \overline B

At stage 2 the output will be \overline {\overline A + \overline B } = A.B

And the final output will be \overline {A.B}

Hence for input A & B the output is \overline {AB} in case of Nand gate.

SSC JE Electrical 2019 with solution SET-1
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If N = linkage flux, then the linkage flux per unit current is defined as:

The inductance L of this coil is defined as the flux linkage per unit current and its unit is Henry.

L =Λ/I

Where

Λ = Flux linkage

I = current

SSC JE Electrical 2019 with solution SET-1
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In the below circuit, X = ?

The circuit shown in the question is an AND GATE. In an AND gate has two or more inputs but it has only one output. An input signal applied to a gate has only two stable states, either 1 (HIGH) or 0 (LOW). In AND gate for any input A&B the output is A.B.