Electric circuits miscellaneous
-  In the circuit shown, i(t) is a unit step current. The steady-state value of v(t) is 
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                        View Hint View Answer Discuss in Forum At steady-state condition, capacitor is open and inductor is shorted, then current through 6 Ω = 1 × 3 = 1 Ω 3 + 6 3 and V(t) = L di + 6i dt = 0 +  6 × 1  = 2 volts 3 
 Correct Option: AAt steady-state condition, capacitor is open and inductor is shorted, then current through 6 Ω = 1 × 3 = 1 Ω 3 + 6 3 and V(t) = L di + 6i dt = 0 +  6 × 1  = 2 volts 3 
 
-  The system function H(s) = 1 s + 1 
 For an input signal cos t, the steady state response is
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                        View Hint View Answer Discuss in Forum Response, C (s) = H (s). R (s) = 1 . s s + 1 s2 + 1 C(t) = - 1 e-t + 1 cos  t - π  2 √2 4 
 At steady state e– t → 0∴ C(t) = 1 .cos  t - π  √2 4 
 Correct Option: AResponse, C (s) = H (s). R (s) = 1 . s s + 1 s2 + 1 C(t) = - 1 e-t + 1 cos  t - π  2 √2 4 
 At steady state e– t → 0∴ C(t) = 1 .cos  t - π  √2 4 
 
-  The steady state in the circuit, shown in the given figure is reached with S open. S is closed at t = 0. The current I at t = 0+ is 
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                        View Hint View Answer Discuss in Forum NA Correct Option: BNA 
-  z-matrix for the network shown in the given figure is 
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                        View Hint View Answer Discuss in Forum From the circuit, 
 V1 = I 1 + (I 1 + I 2). 2s
 = (2s + 1) I 1 + I 2 . 2sand V2 = 132 + (I1 + I2) 2s s = 2s I1 + I2  2s + 3  s  
 Correct Option: AFrom the circuit, 
 V1 = I 1 + (I 1 + I 2). 2s
 = (2s + 1) I 1 + I 2 . 2sand V2 = 132 + (I1 + I2) 2s s = 2s I1 + I2  2s + 3  s  
 
-  Which one of the following parameters does not exist for the two-port network shown in the given figure? 
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                        View Hint View Answer Discuss in Forum Y-parameter = 1  1 -1  Z -1 1 
 and ∆y = 0
 Then Z-parameter can not exist.Correct Option: CY-parameter = 1  1 -1  Z -1 1 
 and ∆y = 0
 Then Z-parameter can not exist.
 
	