Question

When the generator emf in Sample Problem 31.07 is a maximum, what is the voltage across (a) the generator, (b) the resistance, (c) the capacitance, and (d) the inductance? (e) By summing these with appropriate signs, verify that the loop rule is satisfied.

Short answer

1. The voltage across the Generator will be 36.0 V
2. The voltage across the Resistance will be 29.9 V
3. The voltage across the Capacitance will be 11.9 V
4. The voltage across the Inductance will be -5.85 V
5. By summing these with appropriate signs, the loop rule is verified for the given circuit.

Step-by-step solution

Given

$$\begin{gathered} {\xi }_{max}=36.0V \\ f=60.0Hz \\ R=200\Omega \\ L=230mH=230\times {10}^{-3}H \\ C=15.0\mu F=15.0\times {10}^{-6}F \\ l=0.164A \\ \varphi =-24.3^o \end{gathered}$$

Determining the concept

Find the time at which the current is maximum in the circuit. The relationship between the generator voltage and maximum voltage; using this, find the required generator voltage. Having enough information to calculate the current flowing through the resistor. Using Ohm’s law, find the voltage across the resistor. As the impedance, capacitance, and frequency values are given, find the inductive and capacitive reactance. The formula for capacitor voltage and inductor voltage has enough information to calculate the required values.

To verify the loop rule for the circuit, take the summation for voltage across resistor, capacitor, and inductor and compare this with the voltage of the generator. After comparison, conclude whether it is verified or not.

The formulae are as follows:

$$\begin{gathered} \xi ={\xi }_{max}\sin {\omega }_{d}t \\ i=l\sin \left({\omega }_{d}t-\varphi \right) \\ {X}_L=\omega L \\ {X}_C=\frac{1}{\omega C} \\ \omega =2\pi f \\ Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ V=IR \end{gathered}$$

(a) Determining the voltage across the Generator

For,

$$\begin{gathered} \xi ={\xi }_{max} \\ {\omega }_{d}t=\frac{\mathrm{\pi }}{2} \\ t=\frac{\mathrm{\pi }}{2{\omega }_d} \\ t=\frac{1}{4f} \\ t=\frac{1}{4\times 60} \\ t=4.17\times {10}^{-3}s \\ \xi ={\xi }_{max}\sin {\omega }_{d}t \\ \xi ={\xi }_{max}\sin \frac{\mathrm{\pi }}{2} \\ \xi ={\xi }_{max} \\ \xi =36.0V \end{gathered}$$

Hence, the voltage across the Generator will be 36.0 V.

(b) Determining the voltage across the Resistance

$i=l\sin \left({\omega }_{d}t-\varphi \right)$

From the sample problem, l = 0.164 A

$$\begin{gathered} i=0.164\times \sin \left(\frac{\mathrm{\pi }}{2}-\left(-24.3^o\right)\right) \\ i=0.164\times \cos \left(24.3^o\right) \\ i=0.149 \end{gathered}$$

Now,

$$\begin{gathered} V_R=IR \\ {V}_R=0.149\times 200 \\ {V}_R=29.9V \end{gathered}$$

Hence, the voltage across the Resistance will be 29.9 V.

(c) Determining the voltage across the Capacitance

As the capacitor voltage phasor is less than that of the current, so for Capacitor voltage,

$\begin{array}{rcl}{V}_C& =& I{X}_C\sin \left({\omega }_{d}t-\varphi \right)\\ X_C& =& \frac{1}{\omega C}\\ X_C& =& \frac{1}{2\mathrm{\pi fC}}\\ X_C& =& \frac{1}{2\mathrm{\pi }\times 60\times 15.0\times {10}^{-6}}\\ X_C& =& 176.8\\ & \approx & 177\Omega \\ V_C& =& I{X}_C\sin \left({\omega }_{d}t-\varphi \right)\\ V_C& =& 0.164\times 177\times \sin \left(\frac{\mathrm{\pi }}{2}-\left(-24.3^o+\frac{\mathrm{\pi }}{2}\right)\right)\\ V_C& =& 11.9V\end{array}$

Hence, the voltage across the Capacitance will be 11.9 V.

(d) Determining the voltage across the Inductance

As inductor voltage phasor is 90^o. It is more than that of the current, so for Inductor voltage,

$$\begin{gathered} V_L=-I{X}_L\sin \left({\omega }_{d}t-\varphi \right) \\ {X}_L=\omega L \\ {X}_L=2\pi fL \\ {X}_L=2\pi \times 60\times 230\times {10}^{-3} \\ {X}_L=86.7\Omega \\ {V}_L=-0.164\times 86.7\times \sin \left(\frac{\mathrm{\pi }}{2}-\left(-{24}^o+\frac{\mathrm{\pi }}{2}\right)\right) \\ {V}_L=-5.85V \end{gathered}$$

Hence, the voltage across the Inductance will be -5.85 V.

(e) Determining by summing these with appropriate signs, verify that the loop rule is satisfied

$\begin{array}{rcl}\sum V& =& V_L+V_R+V_C\\ \sum V& =& 29.8+11.9+\left(-5.85\right)\\ \sum V& =& 35.85\\ & \approx & 36.0\end{array}$

As the value for the summation of all voltages is equal to the generator voltage, the loop rule is satisfied for the circuit.

Therefore, by summing these with appropriate signs, the loop rule is verified for the given circuit.