Parallel AC Circuit: Impedance, Current and Power Factor
An electrical technician is checking a parallel AC circuit before it is connected to a 20 V rms supply. The two branches have impedances Z₁=6+8i Ω and Z₂=12−5i Ω, where i²=−1 and the supply voltage is taken as 20+0i V rms. The technician's rough note says the combined impedance is 'about 5 Ω' and the rms current is 'about 4 A'. A safety check requires the rms current to stay below 3.2 A and the power factor to remain above 0.95. The technician must calculate the equivalent impedance, the rms current, and the phase angle between voltage and current. Because the branches are in parallel, the equivalent impedance is not the arithmetic mean of the two impedances. A second engineer suggests that a capacitor might be added to correct the power factor, but before any change the existing values must be judged against the safety limits.
1. Calculate the equivalent impedance of the parallel combination in exact and approximate form, and find its modulus.
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For parallel impedances, Z_eq=Z₁Z₂/(Z₁+Z₂). First, Z₁Z₂=(6+8i)(12−5i)=72−30i+96i−40i²=72+66i+40=112+66i. Also Z₁+Z₂=18+3i. Thus Z_eq=(112+66i)/(18+3i). Multiply numerator and denominator by the conjugate 18−3i: numerator (112+66i)(18−3i)=2016−336i+1188i−198i²=2016+852i+198=2214+852i; denominator 18²+3²=324+9=333. So Z_eq=(2214+852i)/333=(738+284i)/111≈6.65+2.56i Ω. Its modulus is |Z_eq|=√((738²+284²)/111²)=√(625300/12321)≈√50.75≈7.12 Ω. The rough note's 'about 5 Ω' is too low.
2. Find the rms current and the phase angle between voltage and current, then compute the power factor.
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The rms current magnitude is I=V/|Z_eq|=20/7.124≈2.81 A. The phase angle of the impedance is arg Z_eq=arctan(284/738)≈21.0°. Since the voltage has phase 0, the current phase is approximately −21.0°; the current lags the voltage by about 21.0°. The power factor is cos(21.0°)≈0.933. This is below the required 0.95. The rms current 2.81 A is below the 3.2 A limit, so the current condition passes, but the power-factor condition fails. The circuit cannot be approved as it stands under the stated safety rule.
3. Assess the rough note's claims of 'about 5 Ω' and 'about 4 A', and state whether the circuit meets both safety limits.
Read the worked answer
The exact modulus is about 7.12 Ω, not 5 Ω, and the exact rms current is about 2.81 A, not 4 A. The rough note underestimates the impedance and overestimates the current, probably by treating the parallel combination too crudely. Against the limits, the current is acceptable because 2.81 A<3.2 A, but the power factor is about 0.933, which is less than 0.95. Since both conditions are required, the circuit does not meet the safety specification. A corrective reactance, such as a capacitor connected in parallel, would be needed to raise the power factor, but the existing values already show that the note's figures should not be used for approval.