Electric circuits miscellaneous
- A certain network N feeds a load resistance R as shown in the Fig. (a) It consumes a power of P W. If an identical network is added as shown in Fig. (b) then power consumed by R will be
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Let network N contains a resistance R and voltage V, then Fig. (a) is redrawn as below.
and P = V 2 = V2 2R 4R
Now, redrawing Fig. (b), we have
power dissipated in R = P = V (2I)2 RP' = 2V 2 R = 4V2 = 16 P 3R 9R 9
Hence, P < P' < 4PCorrect Option: C
Let network N contains a resistance R and voltage V, then Fig. (a) is redrawn as below.
and P = V 2 = V2 2R 4R
Now, redrawing Fig. (b), we have
power dissipated in R = P = V (2I)2 RP' = 2V 2 R = 4V2 = 16 P 3R 9R 9
Hence, P < P' < 4P
- For a 2-port symmetrical bilateral network, if transmission parameters A = 3 and B = 1 Ω, then value of parameter C is
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For symmetrical network
A = D = 3
For bilateral
AD – BC = 1
∴ 9 – C = 1
⇒ C = 8 ΩCorrect Option: B
For symmetrical network
A = D = 3
For bilateral
AD – BC = 1
∴ 9 – C = 1
⇒ C = 8 Ω
- For the network shown below, Z-parameter will be
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Using KVL,
V1(s) = 1 + sL1 I1(s) + SM.I2(s) sC
and V2 (s) = (R + SL2) I 2 (s) + SM I 1 (s)
thenCorrect Option: A
Using KVL,
V1(s) = 1 + sL1 I1(s) + SM.I2(s) sC
and V2 (s) = (R + SL2) I 2 (s) + SM I 1 (s)
then
- For latice type attenuator shown in the given figure, the characteristic impedance R0 is
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Characteristic impedance, R0 = √Rsc Roc
Since Rsc = 2 R1 R2 , Roc = R1 R2 R1 + R2 2
∴ R0 = √R1 R2Correct Option: C
Characteristic impedance, R0 = √Rsc Roc
Since Rsc = 2 R1 R2 , Roc = R1 R2 R1 + R2 2
∴ R0 = √R1 R2
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Y parameters of a four terminal block are 4 2 A single element of 1 ohm 1 1
is connected across as shown in the given figure. The new Y parameters will be
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Y parameter of two networks in parallel are sums of the corresponding Y-parameters.
As Y1 = 1 -1 -1 1 Y2 = 4 2 1 1 ∴ Y = Y1 + Y2 = 5 1 0 2 Correct Option: A
Y parameter of two networks in parallel are sums of the corresponding Y-parameters.
As Y1 = 1 -1 -1 1 Y2 = 4 2 1 1 ∴ Y = Y1 + Y2 = 5 1 0 2