In the circuit shown below, the norton equivalent current in amperes with respect to the terminals P and Q is

6.4 - j4.8

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6.56 - j7.87

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10 + j0

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16 + j0

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Solution

The correct option is A 6.4 - j4.8
For norton equivalent current short circuiting the terminal PQ.

Short circuit current,

$I_{s c} = \frac{25}{15 + j 30 + 25} \times 16 ∠ 0^{0}$

$= \frac{25}{40 + j 30} \times 16 ∠ 0^{0} = \frac{(25 \times 16) ∠ 0^{0}}{50 ∠ 36.86}$

$= 8 ∠ - {36.86}^{0}$

Hence Norton current is

$I_{N} = I_{S C} = 8 ∠ - {36.86}^{0}$

$I_{N} = (6.4 - j 4.8) A$

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In the circuit shown below, the norton equivalent current in amperes with respect to the terminals P and Q is

The correct option is A 6.4 - j4.8For norton equivalent current short circuiting the terminal PQ. Short circuit current, Isc=2515+j30+25×16∠00 =2540+j30×16∠00=(25×16)∠0050∠36.86 =8∠−36.860 Hence Norton current is IN=ISC=8∠−36.860 IN=(6.4−j4.8)A