Chapter 15, Problem 1P

For the resistive element in Fig. 15.81:

1. Write the current in phasor form.
2. Calculate the voltage across the resistor in phasor form.
3. Sketch the phasor diagram of the voltage and current.
4. Write the voltage in the sinusoidal format.
5. Sketch the waveform of the voltage and current.

$i_R=20\times {10}^{-3}\text{sin }\left(1000t+{30}^{\circ }\right)$

Fig. 15.81

To determine

(a)

The Current in phasor form.

Answer to Problem 1P

The current in phasor form is $14.14mA\angle {30}^{\circ }$.

Explanation of Solution

Given:

The sinusoidal expression of current is $20\times {10}^{-3}\sin (1000t+{30}^{\circ })$.

The resistor value is $2k\Omega$.

Concept Used:

In resistive element network, the network consists of only resistor element.

Resistor does not have any phase angle. It's purely real.

In this, the network current is in phase with voltage.

Total impedance $Z=R\angle 0^{\circ }$.

Current: $I=\frac{V}{R}$

  $\begin{array}{l}\text{Where }V\text{ is the voltage}\text{.}\\ R\text{ is the resistance}\text{.}\end{array}$

Calculation:

In order to convert to phasor form

  $\begin{array}{l}\text{Divide by }\sqrt{2}\\ \frac{20\times {10}^{-3}}{\sqrt{2}}\angle {30}^{\circ }\\ 14.14mA\angle {30}^{\circ }\end{array}$

Conclusion:

Hence, the current in phasor form is $14.14mA\angle {30}^{\circ }$.

To determine

(b)

Voltage across resistor in phasor form.

Answer to Problem 1P

Voltage across resistor in phasor form is $28.28V\angle {30}^{\circ }$.

Explanation of Solution

Given:

The sinusoidal expression of current is $20\times {10}^{-3}\sin (1000t+{30}^{\circ })$.

The resistor value is $2k\Omega$.

Concept Used:

In resistive element network, the network consists of only resistor element.

Resistor does not have any phase angle. It's purely real.

In this network current are in phase with voltage.

Total impedance $Z=R\angle 0^{\circ }$.

Current: $I=\frac{V}{R}$

  $\begin{array}{l}\text{Where, }V\text{ is the voltage}\text{.}\\ R\text{ is the resistance}\text{.}\end{array}$

Calculation:

We know that voltage across resistor is given by

  $\begin{array}{l}{\text{v}}_{\text{R}}{\text{=i}}_{\text{R}}\text{R}\\ =14.14mA\angle {30}^{\circ }(2)\\ =28.28V\angle {30}^{\circ }\end{array}$

Conclusion:

Hence, the voltage across resistor in phasor form is $28.28V\angle {30}^{\circ }$.

To determine

(c)

Phasor diagram of the voltage and current.

Answer to Problem 1P

Phasor diagram of the voltage and current is drawn.

  

Explanation of Solution

Given:

The sinusoidal expression of current is $20\times {10}^{-3}\sin (1000t+{30}^{\circ })$.

The resistor value is $2k\Omega$.

Concept Used:

In resistive element network, the network consists of only resistor element.

Resistor does not have any phase angle. It's purely real.

In this, network current are in phase with voltage.

Total impedance $Z=R\angle 0^{\circ }$.

Current: $I=\frac{V}{R}$

  $\begin{array}{l}\text{Where }V\text{ is the voltage}\text{.}\\ R\text{ is the resistance}\text{.}\end{array}$

Calculation:

Based on the voltage and current value which is already obtained in above part the phasor diagram is drawn. From real axis the angle value is counted.

  

Conclusion:

Hence, the phasor diagram of the voltage and current is drawn.

To determine

(d)

Voltage in the sinusoidal expressions.

Answer to Problem 1P

Voltage in the sinusoidal expressions is $v_R=40\sin (1000t+{30}^{\circ })$.

Explanation of Solution

Given:

The sinusoidal expression of current is $20\times {10}^{-3}\sin (1000t+{30}^{\circ })$.

The resistor value is $2k\Omega$.

Concept Used:

In resistive element network, the network consists of only resistor element.

Resistor does not have any phase angle. It's purely real.

In this, the network current is in phase with voltage.

Total impedance $Z=R\angle 0^{\circ }$.

Current: $I=\frac{V}{R}$

  $\begin{array}{l}\text{Where }V\text{ is the voltage}\text{.}\\ R\text{ is the resistance}\text{.}\end{array}$

Calculation:

In order to convert to sinusoidal form,

  $\begin{array}{l}\text{Multiply by }\sqrt{2}\\ v_R=(\sqrt{2})28.28V\angle {30}^{\circ }\\ v_R=40\sin (1000t+{30}^{\circ })\end{array}$

Conclusion:

Hence, voltage in the sinusoidal expressions is $v_R=40\sin (1000t+{30}^{\circ })$.

To determine

(e)

Waveform of the voltage and current.

Answer to Problem 1P

Waveform of the voltage and current is drawn.

  

Explanation of Solution

Given:

The sinusoidal expression of current is $20\times {10}^{-3}\sin (1000t+{30}^{\circ })$.

The resistor value is $2k\Omega$.

Concept Used:

In resistive element network, the network consists of only resistor element.

Resistor does not have any phase angle. It's purely real.

In this, the network current is in phase with voltage.

Total impedance $Z=R\angle 0^{\circ }$.

Current: $I=\frac{V}{R}$

  $\begin{array}{l}\text{Where }V\text{ is the voltage}\text{.}\\ R\text{ is the resistance}\text{.}\end{array}$

Calculation:

Based on the voltage and current value which is already obtained in the above part, the waveform diagram is drawn. Here, A stands for amplitude.

  

Conclusion:

Hence, waveform of the voltage and current is drawn.