Recent content by pgetts

Ah thanks fellas. Finally got the same result as my Mesh analysis. Thanks again.

Thanks guys. So reference node is defined at the bottom. Trivial Nodes would be V10 & V40 as they are actually just V1 and V2 respective So my super node is defined as V20-V30=V3. My currents assumption will be all leaving the supernode so I will get the equation...

Thanks Magoo it's really appreciated as I've been struggling with this for weeks though I'm still not following. As my V20 is connected via the reference node the difference in quantity would be 0-V20 no? or am I missing the point that node V20 would be V1-V20. If you could point me in the...

V1-V20/Z1

Homework Statement [/B] Determine, Using the values given in TABLE A, the current I in the circuit of figure 2 by (a) Mesh analysis (b) Nodal analysis Circuit diagram and given variable attached. V1 = 120 V2 = j120 V3 = 14.142+j14.142 Z1= 2 Z2=-J5 Z3=4 Z4=-J5 Z5=J4Homework Equations [/B] KVL...

Thanks. Yes must make sure that what I input into my calculator is correct. So just to confirm that with an impedance in the form of (a+bi), as the current crosses the impedance this would cause the voltage drop and both the real and imaginary parts sign to change i.e (-a-bi)

Δ = -j4(I1)x(34.999+j41.707)(I2) - (-j6)(I1)x(-34.999-j35.707)(I2) = (381.07-j349.99) ΔI1 = (-j415)x(34.999+j41.707)(I2) - (0) x (-34.999-j35.707)(I2) = (17308.4-j14524.6) ΔI2 = (-j4)(I1)x(0) - (-j6)(I1) x (-j416) = (2490) ∴ I1 = (17308.4-j14524.6)/ (381.07-j349.99) = (3.963+j3.639) I2 =...

O.k so using Cramer's Rule. Δ = -j4(I1)x(34.999+j41.707)(I2) - (-j6)(I1)x(-34.999+j35.707)(I2) = (-47.414 - j349.99) ΔI1 = (-j415)x(34.999+j41.707)(I2) - (0) x (-34.999+j35.707)(I2) = (17308.405-j14524.585) ΔI2 = (-j4)(I1)x(0) - (-j6)(I1) x (-j416) = (2490) I1 = ΔI1/Δ =...

Thanks guys. OK, so here is how I have distributed my currents for voltage source 1. As I move from the top of V1 for loop 1 the first component I encounter is the j4 ohm impedance. As this is with the current flow it would cause a voltage drop so my first component is -j4I1. The next component...

I'm getting I2 for V1 as Δ2/Δ 2490/(-47.41-j350.2) = -0.946+j6.986) I've

1. Homework Statement . Figure 1 shows a 50 Ω load being fed from two voltage sources via their associated reactances. Determine the current i flowing in the load by: (a) Thevenin's theorem (b) Superposition (c) Transforming the two voltage sources and their associated reactances into current...

Hi guys just to say a massive thank you to all the guys and girls out there helping us students out.