Electric circuits miscellaneous
-  With the usual notations, a two-port resistive network satisfies the condition A = D = 3 B = 4 C 2 3 
 z11 of the network is
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                        View Hint View Answer Discuss in Forum Z11 = A = 4 C 4 
 Correct Option: BZ11 = A = 4 C 4 
 
-  A two-port network is defined by the relations I1 = 2V1 + V2, I2 = 2V1 + 3V2 Then z12 is
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                        View Hint View Answer Discuss in Forum From the equations, I1 = 2V1 + (I2 - 2V1) 3 ⇒ V1 = 3I1 - I2 4 ∴ Z12 = 1 Ω 4 
 Correct Option: DFrom the equations, I1 = 2V1 + (I2 - 2V1) 3 ⇒ V1 = 3I1 - I2 4 ∴ Z12 = 1 Ω 4 
 
-  An ideal transformer has turns ratio of 2 : 1. Considering high voltage side as port 1 and low voltage side as port 2, then transmission line parameters of transformer will be
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                        View Hint View Answer Discuss in Forum V1 = I1 = N1 = 2 V2 I2 N2 1 
 ⇒ V1 = 2V2
 I 1 = 0.5 I2 
 In matrix form, V1  =  2 0   V2  I1 0 -0.5 I2 
 Correct Option: CV1 = I1 = N1 = 2 V2 I2 N2 1 
 ⇒ V1 = 2V2
 I 1 = 0.5 I2 
 In matrix form, V1  =  2 0   V2  I1 0 -0.5 I2 
 
-  In a passive two-port network, the open-circuit impedance matrix is  10 2  5 2 
 If input port is interchanged with the output port, then open-circuit impedance matrix will be
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                        View Hint View Answer Discuss in Forum The port-equations are 
 V1 = 10 I 1 + 2I 2 ....(i)
 V2 = 2I 1 + 5 I2 ....(ii)
 Rewriting equations (i) and (ii) as
 V2 = 5 I2 + 2I1 ....(iii)
 V1 = 2 I2 + 10 I1 ....(iv)
 Writing in matrix form V2  =  5 2   I2  V1 2 10 I1 
 Correct Option: BThe port-equations are 
 V1 = 10 I 1 + 2I 2 ....(i)
 V2 = 2I 1 + 5 I2 ....(ii)
 Rewriting equations (i) and (ii) as
 V2 = 5 I2 + 2I1 ....(iii)
 V1 = 2 I2 + 10 I1 ....(iv)
 Writing in matrix form V2  =  5 2   I2  V1 2 10 I1 
 
-  The two electric sub-networks N1 and N2 are connected through three resistors as shown in the figure below. The voltage across 5 Ω resistor and 1 Ω resistor are given to be 10V and 5V respectively. Then voltage across 15 Ω resistor is 
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                        View Hint View Answer Discuss in Forum By KCL, if we take a cutset along all three branches, then total current at the junction is zero. 
 i.e. I 1 + I 2 + I 3 = 0⇒ 10 + I2 + 5 = 0 5 1 
 ⇒ I2 = 7A
 and V = – 105 V.Correct Option: ABy KCL, if we take a cutset along all three branches, then total current at the junction is zero. 
 i.e. I 1 + I 2 + I 3 = 0⇒ 10 + I2 + 5 = 0 5 1 
 ⇒ I2 = 7A
 and V = – 105 V.
 
	