Question

Q2 Two impedances Z1 and 22
are connected in parallel. The
first branch takes a leading
current of 16 A and has a
resistance of 52, while the
second branch takes a lagging
current at a pf of 0.8. The total
power supplied is 5 kW, the
applied voltage being (100 +
j200) V. Determine branch
currents and total current. *
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9:29 PM /

Answer 1

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Answer 2

Given:

I1 = 16 A

R = 52

p.d = 0.8 V

P = 5 kW

V = (100 + j200) V

To find:

1) I2 =?

2) I =?

Explanation:

2)

Power is the product of voltage & current.

P = VI

5000 = (100 + j200) I

I = $\mathbf{10 - j20}$ A

1)

The total current is the sum of current flowing through branches.

     I = I1 + I2

⇒ I2 = I - I1

       = ( ${10 - j20}$ ) - 16

   $\mathbf{I_2 = -6 - j20 \ A}$