Find I in the circuit

  I = 6.364∠3.8° A

Explanation:

You can use KCL or KVL to write node or mesh equations for the voltages and currents in the network. Those require a matrix equation solver capable of working with complex numbers. Some calculators can do that. However, we're going to try a different approach here.

Starting from upper left (j4) and working clockwise around the outside, label the impedances Z1 .. Z4. Label the horizontal branch across the middle Z5. We're going to transform the Δ of Z1, Z2, Z5 into a Y of ZA, ZB, ZC that will facilitate computing the effective impedance of the bridge to the source voltage.

The Δ-Y transformation is symmetrical. The numerator of the equivalent impedance connected to each node is the product of the values currently connected to that node; the denominator is the sum of the values around the loop of the Δ.

So, If we transform the Δ of Z1, Z2, Z5 to a Y of ZA, ZB. ZC with ZA connected where Z1 and Z2 are now connected, ZB connected to Z4, and ZC connected to Z3, the network becomes a series-parallel network with an effective impedance of ...

  Z = ZA + ((ZB +Z4) ║ (ZC +Z3))

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For starters, we have ...

  ZA = Z1·Z2/(Z1 +Z2 +Z5) = (j4)(-j3)/(j4-j3+8+j5) = 12/(8+j6) = 0.96-j0.72

  ZB = (j4)(8+j5)/(8+j6) = (20+j32)/(8+j6) = 0.32+j3.76

  ZC = (-j3)(8+j5)/(8+j6) = (15-j24)/(8+j6) = -0.24-j2.82

So, the left branch of the parallel combination is ...

  ZB +Z4 = (0.32+j3.76) +(5-j2) = 5.32+j1.76

And the right branch is ...

  ZC +Z3 = (-0.24-j2.82) +10 = 9.76-j2.82

Then the series-parallel combination we want is ...

  ZA + (ZB+Z4)(ZC+Z3)/(ZB+Z4+ZC+Z3) ≈ 4.703671 -j0.3126067

That is, the impedance of the bridge circuit to the source voltage is about ...

  4.7140478∠-3.802°

Dividing the source voltage by this impedance gives the source current, ...

  I = (30∠0°)/(4.7140478∠-3.802°)

  I ≈ 6.363958∠3.802° . . . amperes

answer:

a

explanation:

answer:

jiggy jiggy jab my titty

explanation: