Electric circuits miscellaneous
- In the circuit shown in the given figure, the values of I(0+) and I(∞), will be, respectively
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I(s) = 3 (s + 1) 1 + 1 2s = 3 (s + 0.5)(s + 1) = -3 + 6 (s + 0.5) (s + 1)
⇒ i (t) = L– 1(I(s)) = 6e– t – 3e– 0.5t
At t = 0, i (0+) = 3A,
and at t = ∞, i (∞) = 0Correct Option: C
I(s) = 3 (s + 1) 1 + 1 2s = 3 (s + 0.5)(s + 1) = -3 + 6 (s + 0.5) (s + 1)
⇒ i (t) = L– 1(I(s)) = 6e– t – 3e– 0.5t
At t = 0, i (0+) = 3A,
and at t = ∞, i (∞) = 0
- Consider the graph and tree (dotted) of the given figure. The fundamental loops include the set of lines
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NA
Correct Option: D
NA
- A connected network of N > 2 nodes has at most one branch directly connecting any pair of nodes. The graph of the network
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NA
Correct Option: D
NA
- The number of independent loops for a network with n nodes and b branch is
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NA
Correct Option: C
NA
- In the circuit shown in the figure, X is an element which always absorbs power during a particular operation it sets up a current of 1 amp in the direction shown and absorbs a power Px. It is possible that X can absorb the same power Px for another current I. the value of this current is
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As element absorbs power, let it be Rx.
i = 1 = 6 Rx + 1
⇒ Rx = 5 ΩP = 6 2 × 5 = 5 W 6
For 5A, P = 52 × 5 = 125 W
For 3 + √14, P = 227 W
For 3 – √14, P = 2.75 W
Hence none of the above is correct.Correct Option: D
As element absorbs power, let it be Rx.
i = 1 = 6 Rx + 1
⇒ Rx = 5 ΩP = 6 2 × 5 = 5 W 6
For 5A, P = 52 × 5 = 125 W
For 3 + √14, P = 227 W
For 3 – √14, P = 2.75 W
Hence none of the above is correct.