Question

a) A p.d. of 200∠30° V is applied to two branches connected in parallel. The currents in the respective branches are 20∠60° A and 40∠30° A. Find the apparent power (in kV A) and the active power (in kW) in each branch and in the main network.

Answer 1

Given that,

Potential difference = 200 V

Angle = 30°

Current in first branches= 20 A

Angle = 60°

Current in second branches= 40 A

Angle = 30°

We need to calculate the apparent power in branch first

Using formula of power

$P_{1}=V\times I_{1}$

Put the value in to the formula

$P_{1}=200\times 20$

$P_{1}=4000\ kVA$

We need to calculate the apparent power in branch second

Using formula of power

$P_{1}=V\times I_{2}$

Put the value in to the formula

$P_{1}=200\times 40$

$P_{1}=8000\ kVA$

We need to calculate the phase difference between voltage and current in first branch

Using formula of phase difference

$\phi=\phi_{2}-\phi_{1}$

$\phi=60-30$

$\phi=30^{\circ}$

We need to calculate the power in first branch

Using formula of power

$P_{1}=VI_{1}\cos\phi_{1}$

Put the value into the formula

$P_{1}=200\times20\cos30$

$P_{1}=3464\W$

$P_{1}=3.5\ kW$

The phase difference between voltage and current in second branch is zero.

We need to calculate  the power in second branch

Using formula of power

$P_{2}=VI\cos\phi_{2}$

Put the value into the formula

$P_{2}=200\times40\cos0$

$P_{2}=8000\ W$

$P_{2}=8\ kW$

The current I₁ in a complex form

$I_{1}=20(\cos60+j\sin60)$

$I_{1}=20(0.5+j0.86)$

$I_{1}=10+j17.2$

The current I₂ in a complex form

$I_{2}=40(\cos30+j\sin30)$

$I_{2}=40(0.86+j0.5)$

$I_{2}=34.4+j20$

We need to calculate the total current

Using formula of total current

$I=I_{1}+I_{2}$

Put the value into the formula

$I=10+j17.2+34.4+j20$

$I=44.4+j(17.2+20)$

$I=44.4+j(37.2)$

$I=\sqrt{(44.4)^2+(37.2)^2}$

$I=58\ A$

We need to calculate the direction of current

Using formula of direction

$\tan\phi=\dfrac{j}{i}$

Put the value into the formula

$\phi=\tan^{-1}(\dfrac{37.2}{44.4})$

$\phi=39.9^{\circ}$

We need to calculate the total apparent power in the circuit

Using formula of power

$P=VI$

Put the value into the formula

$P=200\times58$

$P=11600\ watt$

We need to calculate the active power in circuit

Using formula of active power

$P'=VI\cos\phi$

Put the value into the formula

$P'=200\times58\cos(39.8-30)$

$P'=11430.7\ Watt$

$P'=11.43\ kW$

Hence, The total apparent power in the circuit is 11.6 kW.

The active power in circuit is 11.43 kW.