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Test: Resistance & Inductance in Series


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10 Questions MCQ Test Basic Electrical Technology | Test: Resistance & Inductance in Series

Test: Resistance & Inductance in Series for Electrical Engineering (EE) 2022 is part of Basic Electrical Technology preparation. The Test: Resistance & Inductance in Series questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Resistance & Inductance in Series MCQs are made for Electrical Engineering (EE) 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Resistance & Inductance in Series below.
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Test: Resistance & Inductance in Series - Question 1

A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.

Detailed Solution for Test: Resistance & Inductance in Series - Question 1

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.

Test: Resistance & Inductance in Series - Question 2

 A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the phase difference.

Detailed Solution for Test: Resistance & Inductance in Series - Question 2

 φ=tan-1(XL/R)=55.1
Since this is an inductive circuit, the current will lag, hence φ= -55.1.

Test: Resistance & Inductance in Series - Question 3

A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the resistor.

Detailed Solution for Test: Resistance & Inductance in Series - Question 3

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.
Voltage across resistor= 8.2*7=57.4V.

Test: Resistance & Inductance in Series - Question 4

A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the inductor.

Detailed Solution for Test: Resistance & Inductance in Series - Question 4

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.
Voltage across resistor= 8.2*7=57.4V.
Voltage across inductor =100-VR= 42.6V.

Test: Resistance & Inductance in Series - Question 5

A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a x V 50Hz sinusoidal supply. The current in the circuit is 8.2A. Calculate the value of x.

Detailed Solution for Test: Resistance & Inductance in Series - Question 5

 XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, Therefore V=12.2*8.2=100V.

Test: Resistance & Inductance in Series - Question 6

Which, among the following, is the correct expression for φ.

Detailed Solution for Test: Resistance & Inductance in Series - Question 6

Form the impedance triangle, we get tanφ= XL/R.
Hence φ=tan-1 (XL/R).

Test: Resistance & Inductance in Series - Question 7

For an RL circuit, the phase angle is always ________

Detailed Solution for Test: Resistance & Inductance in Series - Question 7

For a series resistance and inductance circuit the phase angle is always a negative value because the current will always lag the voltage.

Test: Resistance & Inductance in Series - Question 8

 What is φ in terms of voltage?

Detailed Solution for Test: Resistance & Inductance in Series - Question 8

Form the voltage triangle, we get cosφ= VR/V.
Hence φ=cos-1VR/V.

Test: Resistance & Inductance in Series - Question 9

An RL network is one which consists of?

Detailed Solution for Test: Resistance & Inductance in Series - Question 9

An R-L network is a network which consists of a resistor which is connected in series to an inductor.

Test: Resistance & Inductance in Series - Question 10

At DC, inductor acts as ___________

Detailed Solution for Test: Resistance & Inductance in Series - Question 10

At DC, the inductor acts as short circuit because the inductive resistance is zero. The frequency of a DC circuit is 0. The inductive resistance=(2*pi*f*L). Therefore, if the frequency is 0, the inductive resistance is zero and it acts as an short circuit.

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