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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                        View Hint View Answer Discuss in Forum 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: CI(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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                        View Hint View Answer Discuss in Forum NA Correct Option: DNA 
-  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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                        View Hint View Answer Discuss in Forum NA Correct Option: DNA 
-  The number of independent loops for a network with n nodes and b branch is
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                        View Hint View Answer Discuss in Forum NA Correct Option: CNA 
-  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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                        View Hint View Answer Discuss in Forum 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: DAs 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.
 
	