Problem

# For the network in Fig. 1:a. Find the current I.b. Find the voltage VCc. Find the average...

For the network in Fig. 1:

a. Find the current I.

b. Find the voltage VC

c. Find the average power delivered to the network.

FIG. 1

#### Step-by-Step Solution

Solution 1

Given circuit diagram:

(a)

Consider the following network to find the current:

In the above network, impedance $$\mathbf{Z}_{1}$$ is given by

\begin{aligned} \mathbf{Z}_{1} &=j X_{L}-j X_{C_{1}} \\ &=j 600 \Omega-j 200 \Omega \\ &=j 400 \Omega \\ \mathbf{Z}_{1} &=400 \Omega \angle 90^{\circ} \end{aligned}

Now current $$\mathbf{I}_{1}$$ is given by

\begin{aligned} \mathbf{I}_{1} &=\frac{\mathbf{E}}{\mathbf{Z}_{1}} \\ &=\frac{100 \mathrm{~V} \angle 0^{\circ}}{400 \Omega \angle 90^{\circ}} \\ \mathbf{I}_{1} &=0.25 \mathrm{~A} \angle-90^{\circ} \end{aligned}

In the above network, impedance $$\mathbf{Z}_{2}$$ is given by

\begin{aligned} \mathbf{Z}_{2} &=R_{2}+\left(X_{C_{1}} \| X_{C_{2}}\right) \\ &=100 \Omega+\frac{\left(400 \Omega \angle-90^{\circ}\right)\left(400 \Omega \angle-90^{\circ}\right)}{\left(400 \Omega \angle-90^{\circ}\right)+\left(400 \Omega \angle-90^{\circ}\right)} \\ &=100 \Omega+200 \Omega \angle-90^{\circ} \\ \mathbf{Z}_{2} &=223.61 \angle-63.43^{\circ} \Omega \end{aligned}

Now current $$\mathbf{I}_{2}$$ is given by

\begin{aligned} \mathbf{I}_{2} &=\frac{\mathbf{E}}{\mathbf{Z}_{2}} \\ &=\frac{100 \mathrm{~V} \angle 0^{\circ}}{223.61 \Omega \angle-63.43^{\circ}} \\ \mathbf{I}_{2} &=0.447 \angle 63.43^{\circ} \mathrm{A} \end{aligned}

Current I is given by

\begin{aligned} \mathbf{I} &=\mathbf{I}_{1}+\mathbf{I}_{2} \\ &=0.25 \mathrm{~A} \angle-90^{\circ}+0.447 \mathrm{~A} \angle 63.43^{\circ} \\ \mathbf{I} &=0.25 \mathrm{~A} \angle 36.86^{\circ} \end{aligned}

Therefore, the total current $$\mathbf{I}$$ in the network is $$\mathbf{I}=0.25 \angle 36.86^{\circ} \mathrm{A}$$.

(b)

From the given network, voltage $$\mathbf{V}_{C}$$ is given by

\begin{aligned} \mathbf{V}_{C} &=\left(X_{C_{1}} \| X_{C_{2}}\right)\left(\mathbf{I}_{2}\right) \\ &=\frac{\left(400 \Omega \angle-90^{\circ}\right)\left(400 \Omega \angle-90^{\circ}\right)}{\left(400 \Omega \angle-90^{\circ}\right)+\left(400 \Omega \angle-90^{\circ}\right)}\left(0.447 \mathrm{~A} \angle 63.43^{\circ}\right) \\ &=\left(200 \Omega \angle-90^{\circ}\right)\left(0.447 \mathrm{~A} \angle 63.43^{\circ}\right) \\ \mathbf{V}_{C} &=89.4 \angle-26.57^{\circ} \end{aligned}

Therefore, the voltage $$\mathbf{V}_{C}$$ in the network is $$\mathbf{V}_{C}=89.4 \angle-26.57^{\circ} \mathrm{V}$$.

(c)

The average power delivered to the network is given by

\begin{aligned} P_{I} &=E I \cos \left(\theta_{T}\right) \\ &=(100 \mathrm{~V})(0.25 \mathrm{~A}) \cos (36.86) \\ P_{T} &=20 \mathrm{~W} \end{aligned}

Therefore, the average power delivered to the given network is $$P_{T}=20 \mathrm{~W}$$.