Showing posts with label Network Analysis. Show all posts
Showing posts with label Network Analysis. Show all posts
Driving point impedance
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The above property will be satisfied by
(a)R L network only
(b)RC network only
(c)LC network only
(d)RC & R L networks
Answer is ( )
Concept
Type Reasoning here
Current Flow through Non - Uniform Area
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(a) current will be different at different cross-sections.
(b) current will be the same at all the cross-sections.
(c) current will be different but current density will be same at all the cross-sections
(d) current will be the same but current density will be different at different cross-sections
Answer is ( )
Concept
Type Reasoning here
Maximum Load Transfer
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A source of angular frequency 1 rad/ s has a source impedance consisting of 1 ohm resistance in series with 1H inductance.The load that will obtain the maximum transfer is
(a)1 Ohm
(b) 1 Ohm Resistance in Parallel with 1 H inductance
(c) 1 Ohm Resistance in series with 1 F Capacitor
(d) 1 Ohm Resistance in Parallel with 1 F Capacitor
Answer is ( )
Concept
Type Reasoning here
Average Power - Passive Elements
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A 10mH inductor carries a sinusoidal current of 1A rms at a frequency of 50Hz.The average power dissipated by the inductor is
(a) 0 W
(b) 0.25 W
(c) 0.5 W
(d) 1 W
Answer is ( a )
Concept
Explanation:The average power dissipated by a circuit which has only reactive element(purely capacitive or purely inductive) is 0.
Active Power
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An AC Source of 200V rms supplies an active power of 600W and reactive power of 800VAR. The rms current drawn from the source is
(a) 10A
(b) 5A
(c) 3.75A
(d) 2.5A
Concept
Explanation: $V_{s} = 200V (rms)$$P = 600W , Q = 800VAR$$P = V_{s}I_{s}cos\phi = 600$$\therefore I_{s}cos\phi = 600/200 = 3 $...Eqn(1)$Q = V_{s}I_{s}sin\phi = 800$$\therefore I_{s}sin\phi = 800/200 = 4 $...Eqn(2)$\therefore$ From Eqn(1) and (2) $I_{s}^{2}(sin^{2}\phi + cos^{2}\phi)$ = 9 +16 = 25
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