Network theory miscellaneous Moderate Questions and Answers | Page - 15

Network theory miscellaneous

For the network shown below, the power consumed in the resistance R is measured when one source is acting at a time, these values are 18W, 50W and 98W. When all the sources are acting simultaneously, the possible maximum and minimum values of power in R will be—

1. 98W and 18W
1. 166W and 18W
1. 450W and 2W
1. 166W and 2W

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The given circuit:

This problem can be solved by using superposition theorem.
Case 1: When source E₁ is taken:
P₁ = 18 = E²₁/R
Let R = 2Ω:
E²₁ = 18 × 2 = 36
or
E₁ = 6V
Case 2:
When source E₂ is taken:
P₂ = 50 = E²₂/R
or
E²₂ = 50 × 2 = 100
or
E₂ = 10V
Case 3: When source E₃ is taken:
P₃ = 98 = E³₂/R
or
E²₃ = 98 × 2 = 196
or
E₃ = 14V
When all the sources are acting: Then maximum value of voltage,
= E_max = E₁ + E₂ + E₃ = 6 + 10 + 14 = 30V
P_max = E²_max = 30² = 450W
R 2

And minimum value of voltage,
E_min = E₁ + E₂ – E₃ = 6 + 10 – 14 = 2V
P_min = E²_min = 2² = 2W
R 2

Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit:

This problem can be solved by using superposition theorem.
Case 1: When source E₁ is taken:
P₁ = 18 = E²₁/R
Let R = 2Ω:
E²₁ = 18 × 2 = 36
or
E₁ = 6V
Case 2:
When source E₂ is taken:
P₂ = 50 = E²₂/R
or
E²₂ = 50 × 2 = 100
or
E₂ = 10V
Case 3: When source E₃ is taken:
P₃ = 98 = E³₂/R
or
E²₃ = 98 × 2 = 196
or
E₃ = 14V
When all the sources are acting: Then maximum value of voltage,
= E_max = E₁ + E₂ + E₃ = 6 + 10 + 14 = 30V
P_max = E²_max = 30² = 450W
R 2

And minimum value of voltage,
E_min = E₁ + E₂ – E₃ = 6 + 10 – 14 = 2V
P_min = E²_min = 2² = 2W
R 2

Hence alternative (C) is the correct choice.