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It's parallel, can you tell me why it computes like serial?
second answer
For resistors in series, R = R1 + R2 + R3 +...
In parallel, 1/R = 1/R1 + 1/R2 +1/R3 + ....
An inductor is like a resistor.
For capacitors in series, 1/C = 1/C1 + /1C2 + 1/C3 + ....
In parallel, C = C1 + C2 + C3 + .....
I think you're confused because the calculations for capacitors and resistors are opposite to each other.
The explanation in the above textbook is wrong.
If -j3Ω and -j7Ω are connected in parallel, of course -j3Ω || -j7Ω = -j2.1Ω should come out.
This cannot be -j10Ω.
Let's prove this.
1/(jωC1) = -j3 so ωC1 = 1/3, and 1/(jωC2) = -j7 so ωC2 = 1/7.
Then, combining these capacities, ωCT = ωC1+ωC2 = 1/3 + 1/7 = 10/21,
ZT = 1/(jωCT) = -j2.1Ω.
second answer
For resistors in series, R = R1 + R2 + R3 +...
In parallel, 1/R = 1/R1 + 1/R2 +1/R3 + ....
An inductor is like a resistor.
For capacitors in series, 1/C = 1/C1 + /1C2 + 1/C3 + ....
In parallel, C = C1 + C2 + C3 + .....
I think you're confused because the calculations for capacitors and resistors are opposite to each other.
The explanation in the above textbook is wrong.
If -j3Ω and -j7Ω are connected in parallel, of course -j3Ω || -j7Ω = -j2.1Ω should come out.
This cannot be -j10Ω.
Let's prove this.
1/(jωC1) = -j3 so ωC1 = 1/3, and 1/(jωC2) = -j7 so ωC2 = 1/7.
Then, combining these capacities, ωCT = ωC1+ωC2 = 1/3 + 1/7 = 10/21,
ZT = 1/(jωCT) = -j2.1Ω.