Two impedances \(Z_1=(6+j8)\ \Omega\) and \(Z_2=(8-j6)\ \Omega\) are connected in parallel across a 200 V, 50 Hz single-phase supply. Find the admittance and current of each branch, the total current, the overall power factor and the total power.

Answer: \(Y_1=0.06-j0.08\) S, \(Y_2=0.08+j0.06\) S; \(I_1=20\angle-53.13^\circ\) A, \(I_2=20\angle36.87^\circ\) A; \(I=28.28\angle-8.13^\circ\) A; pf \(=0.99\) lagging; \(P=5600\) W.

Admittances: \(Y_1=\dfrac{1}{6+j8}=\dfrac{6-j8}{100}=0.06-j0.08\) S; \(Y_2=\dfrac1{8-j6}=\dfrac{8+j6}{100}=0.08+j0.06\) S.

Total \(Y=Y_1+Y_2=0.14-j0.02\) S \(=0.1414\angle-8.13^\circ\) S.

Branch currents (take \(V=200\angle0^\circ\)):

\(I_1=VY_1=12-j16=20\angle-53.13^\circ\) A, pf \(=\cos53.13^\circ=0.6\) lagging.

\(I_2=VY_2=16+j12=20\angle36.87^\circ\) A, pf \(=0.8\) leading.

Total current: \(I=I_1+I_2=28-j4=28.28\angle-8.13^\circ\) A.

Power factor: \(\cos8.13^\circ=0.99\) lagging.

Power: \(P=VI\cos\phi=200\times28.28\times0.99=5600\) W. Check: \(I_1^2R_1+I_2^2R_2=400\times6+400\times8=5600\) W ✓.