1
GATE ECE 2011
+2
-0.6
In the circuit shown below, the network N is described by the following Y matrix: $$Y\;=\;\begin{bmatrix}0.1S&-0.01S\\0.01S&0.1S\end{bmatrix}$$\$ The voltage gain $$\frac{V_2}{V_1}$$ is
A
1/90
B
-1/90
C
-1/99
D
-1/11
2
GATE ECE 2008
+2
-0.6
A two-port network shown below is excited by external dc sources. The voltages and the currents are measured with voltmeters $${V_1}$$, $${V_2}$$ and ammeter $${A_1}$$, $${A_2}$$ (all assumed to be ideal), as indicated. Under following switch conditions, the readings obtained are:

(i) $${S_1} - open,\,\,\,\,{S_2} - closed$$
$${A_1} = 0\,A,\,\,\,\,\,\,{V_1} = \,4.5\,\,V,$$
$${V_2}\, = \,1.5\,V,\,\,\,\,{A_2}\, = \,1\,A$$

(ii) $${S_1} - Closed,\,\,\,\,{S_2} - Open$$
$${A_1} = 4\,A,\,\,\,\,\,\,{V_1} = \,6\,\,V,$$
$${V_2}\, = \,6\,V,\,\,\,\,{A_2}\, = \,0\,A$$

The z-parameter matrix for this network is
A
$$\left[ {\matrix{ {1.5} & {1.5} \cr {4.5} & {1.5} \cr } } \right]$$
B
$$\left[ {\matrix{ {1.5} & {4.5} \cr {1.5} & {4.5} \cr } } \right]$$
C
$$\left[ {\matrix{ {1.5} & {4.5} \cr {1.5} & {1.5} \cr } } \right]$$
D
$$\left[ {\matrix{ {4.5} & {1.5} \cr {1.5} & {4.5} \cr } } \right]$$
3
GATE ECE 2008
+2
-0.6
A two-port network shown below is excited by external dc sources. The voltages and the currents are measured with voltmeters $${V_1}$$, $${V_2}$$ and ammeter $${A_1}$$, $${A_2}$$ (all assumed to be ideal), as indicated. Under following switch conditions, the readings obtained are:

(i) $${S_1} - open,\,\,\,\,{S_2} - closed$$
$${A_1} = 0\,A,\,\,\,\,\,\,{V_1} = \,4.5\,\,V,$$
$${V_2}\, = \,1.5\,V,\,\,\,\,{A_2}\, = \,1\,A$$

(ii) $${S_1} - Closed,\,\,\,\,{S_2} - Open$$
$${A_1} = 4\,A,\,\,\,\,\,\,{V_1} = \,6\,\,V,$$
$${V_2}\, = \,6\,V,\,\,\,\,{A_2}\, = \,0\,A$$

The h-parameter matrix for this network is
A
$$\left[ {\matrix{ { - 3} & 3 \cr { - 1} & {0.67} \cr } } \right]$$
B
$$\left[ {\matrix{ { - 3} & -1 \cr { 3} & {0.67} \cr } } \right]$$
C
$$\left[ {\matrix{ { 3} & 3 \cr { 1} & {0.67} \cr } } \right]$$
D
$$\left[ {\matrix{ { 3} & 1 \cr { - 3} & {-0.67} \cr } } \right]$$
4
GATE ECE 2008
+2
-0.6
The driving point impedance of the following network is given by $$Z(s) = {{0.2\,s} \over {{s^2}\, + \,0.1\,s\, + \,2}}$$. The component values are
A
L = 5H, R = 0.5 $$\Omega$$, C = 0.1 F
B
L = 0.1 H, R = 0.5 $$\Omega$$, C = 5 F
C
L = 5H, R = 2 $$\Omega$$, C = 0.1 F
D
L = 0.1 H, R = 2 $$\Omega$$, C = 5 F
EXAM MAP
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