Network Elements and the Concept of Circuit
- In a balanced Wheatstone bridge, if the positions of detector and source are interchanged, the bridge will still remain balanced. This inference can be drawn from—
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NA
Correct Option: A
NA
- The value of the resistance R in the circuit shown in the given figure is varied in such a manner that the power dissipated in 5 Ω resistor is maximum. Under this condition, the value of R will be—
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Power draw through the 5 Ω (i.e. load resistance) will be maximum if and only if Rth = RL calculation for Rth for the given circuit.
Rth = 8 R 8 + R 8 R = 5 8 + R
8R = 5R + 40
3R = 40R = 40 Ω 3 Correct Option: B
Power draw through the 5 Ω (i.e. load resistance) will be maximum if and only if Rth = RL calculation for Rth for the given circuit.
Rth = 8 R 8 + R 8 R = 5 8 + R
8R = 5R + 40
3R = 40R = 40 Ω 3
- Calculate the Thevenin resistance for the circuit shown below between A and B—
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Calculation of Rth:
Rth = 10 + (2 + 2) || 4= 10 + 4 × 4 4 + 4
= 10 + 2 = 12 ΩCorrect Option: B
Calculation of Rth:
Rth = 10 + (2 + 2) || 4= 10 + 4 × 4 4 + 4
= 10 + 2 = 12 Ω
- The rms voltage measured across on admittance (G + jB) is V. The reactive power for the element is—
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NA
Correct Option: A
NA
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, then the value of lim t→∞ f (t)—If L {f (t)} = s2 + 2
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Given, L {f (t)} = s2 + 2
or f (t) = sin t.
as t → ∞, f (t) = sin t goes through 1 to 0
and – 1 to 0 to infinite times so the value of Lim t→∞ f(t) cannot be determined.Correct Option: A
Given, L {f (t)} = s2 + 2
or f (t) = sin t.
as t → ∞, f (t) = sin t goes through 1 to 0
and – 1 to 0 to infinite times so the value of Lim t→∞ f(t) cannot be determined.