Question

For the single phase circuit shown in figure 2.22,l=10 ∠0^oa) Compute the phasors, l_{1 ,}l₂and V(b) Draw a phasor diagram showing,l_{1 ,}l₂and V.

Short answer

(a) The Phasor of the current l₁is 7.5 ∠-90^oA , phasor of the current l₂ 12.5∠ 36.87^oAand the voltage phasor V is 75 ∠-53.13^oV.

(b) The phasor diagram representing, and .l_{1 ,}l₂and V

Step-by-step solution

Write the given data of the question

The given circuit consists of resistance, inductance and capacitance whose value is given as;

The value of the Resistance is given by,

R = $\mathrm{\Omega }$

The value of thecapacitive reactance is given by,

X_{c = j6$\mathrm{\Omega }$}

_{value of the Inductive reactance is given by,}

X_L= j6$\mathrm{\Omega }$

The value for the Supply current is given by,

l = 10∠ 0^oA

Write the concept for polar form and rectangular form:

If x (t) =Acoswt is taken as the reference for the phasor diagram.

The polar form or phasor expression of y (t) = Acos (wt-θ) is,

Here, means that y (t) lags x (t) by θ angle, A is the maximum or peak value of y (t) and i$\frac{A}{\sqrt[]{2}}$is the rms value of the y (t).

The phasor diagram is shown below;

Calculate phasor of the current and current l2:

(a)

Apply the current division rule to find the phasor of current l₁.

Substitute 10∠0^o for l in the above equation.

Therefore, the phasor of the current l₁is 7.5∠ -90^o.

Again, Apply the current division rule to find the phasor of current l₂.

Substitute 10∠ -0^o. for l in the above equation.

Therefore the phasor of the current l₁is 12.5∠ 36.87^o

Calculate the phasor expression for the voltage V.

(b)

Since the given circuit is a parallel circuit, the same voltage source is applied across both parallel branches. So calculate the voltage drop across the either of parallel branch.

V = (8+j6)i₁

Substitute7.5∠ -90^o. for l₁in the equation.

Therefore the phasor for the voltage V is 75∠ -53.13^oV.

The phasor diagram representing, ,l_{1 ,}l₂ and V is drawn below