Electric circuits miscellaneous Easy Questions and Answers | Page - 33

Electric circuits miscellaneous

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of

1. k²
1. k
1. 1 / k
1. √k

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R_c = R_a.R_b as R_a is scaled by facts k
R_a + R_b + R_c

R'_c = R'_a.R'_b
R'_a + R_b + R_c

= k²R_a.R_b = k. R_a × R_b
k(R_a + R_b + R_c) R_a + R_b + c

so elements corresponding to star equivalence will be seated by facts k.

Correct Option: B

R_c = R_a.R_b as R_a is scaled by facts k
R_a + R_b + R_c

R'_c = R'_a.R'_b
R'_a + R_b + R_c

= k²R_a.R_b = k. R_a × R_b
k(R_a + R_b + R_c) R_a + R_b + c

so elements corresponding to star equivalence will be seated by facts k.

In the circuit shown below, if the source voltage V_s =100 ∠ 53.13° V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance R_L is

1. 100 ∠90°
1. 800 ∠0°
1. 800 ∠90°
1. 100 ∠60°

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Given : V_s = 100 ∠53.13° V

To find : Thevenin’s voltage across Load resistonce
Solution
* For V'_th open it.

* Opening, then I₂ = 0
* When I₂ = 0, then j40I₂ = 0 {voltage source will short circuit)
∴ Circuit became

* V_TH = 10V_L1 because no-current flowing through circuit.
V_TH = 10 × j4 × V_s = 40 j V_s
3 + 4j 3 + 4j

From rectangular domain to polar domain.

Correct Option: C

Given : V_s = 100 ∠53.13° V

To find : Thevenin’s voltage across Load resistonce
Solution
* For V'_th open it.

* Opening, then I₂ = 0
* When I₂ = 0, then j40I₂ = 0 {voltage source will short circuit)
∴ Circuit became

* V_TH = 10V_L1 because no-current flowing through circuit.
V_TH = 10 × j4 × V_s = 40 j V_s
3 + 4j 3 + 4j

From rectangular domain to polar domain.

In the circuit shown below, the current through the inductor is

1. 2 A
  1 + j
1. -1 A
  1 + j
1. 1 A
  1 + j
1. 0 A

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I_L = 1∟0 × 1 = 1 A
1 + j1 1 + j1

Correct Option: C

I_L = 1∟0 × 1 = 1 A
1 + j1 1 + j1

The impedance looking into nodes 1 and 2 in the given circuit is

1. 50 Ω
1. 100 Ω
1. 5 kΩ
1. 10.1 kΩ

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After connecting a voltage source of V
R_th = V = 50I = 50 Ω
I I

V₁ = V₂
⇒ (10k) (– i_b) = 100 (I + 99i_b + i_b)
and – 10000i_b = 100I + 100 × 100i_b
= 100I + 10000i_b
⇒ – 20000i_b = 100I
i_b = - 100 I = - I
20000 200

= 50I
⇒ V/I = 50 Ω

Correct Option: A

After connecting a voltage source of V
R_th = V = 50I = 50 Ω
I I

V₁ = V₂
⇒ (10k) (– i_b) = 100 (I + 99i_b + i_b)
and – 10000i_b = 100I + 100 × 100i_b
= 100I + 10000i_b
⇒ – 20000i_b = 100I
i_b = - 100 I = - I
20000 200

= 50I
⇒ V/I = 50 Ω

The bus admittance matrix of a three-bus three-line system is

If each transmission line between the two buses is represented by an equivalent π-network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is

1. 4
1. 2
1. 1
1. 0

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For three bus system

Y₁₁ = Y₁₁ + Y₁₂ + Y₁₃
Y₁₂ = – Y₁₂ = 10j, Y₁₃ = – Y₁₃ = 5j
⇒ Y₁₁ = – 13j = Y₁₁ + Y₁₂ + Y₁₃
⇒ – 13j = Y₁₁ – 10j – 5j
⇒ Y₁₁ = 2j
Equivalent π network between Bus 1 and Bus 2

Shunt-susceptance, y₁₁ = 2

Correct Option: B

For three bus system

Y₁₁ = Y₁₁ + Y₁₂ + Y₁₃
Y₁₂ = – Y₁₂ = 10j, Y₁₃ = – Y₁₃ = 5j
⇒ Y₁₁ = – 13j = Y₁₁ + Y₁₂ + Y₁₃
⇒ – 13j = Y₁₁ – 10j – 5j
⇒ Y₁₁ = 2j
Equivalent π network between Bus 1 and Bus 2

Shunt-susceptance, y₁₁ = 2