Chapter 21, Problem 24P

An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an inductor (L = 2.8 H). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the inductor? (c) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the inductor, and the AC source? (d) When the current is zero, what are the magnitudes of the potential difference across the resistor, the inductor, and the AC source?

To determine

The maximum current in the circuit.

Answer to Problem 24P

The maximum current in the circuit is $0.11\text{A}$.

Explanation of Solution

Given Info: The resistance of the resistor is $1.2\text{kΩ}$. The inductance of the inductor is $2.8\text{H}$. The frequency of the ac circuit is $60\text{Hz}$. Maximum voltage of the source is $170\text{V}$.

The inductive reactance of the circuit is,

$X_L=2\pi fL$

• $f$ is the frequency of the ac source
• $L$ is the inductance of the inductor

Substitute $60\text{Hz}$ for $f$, $2.8\text{H}$ for $L$ to determine the inductive reactance,

$\begin{array}{c}{X}_L=2\pi \left(60\text{Hz}\right)\left(2.8\text{H}\right)\\ =1.1\times {10}^3\text{Ω}\end{array}$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

• $R$ is the resistance of the resistor
• $X_L$ is the inductive reactance
• $X_C$ is the capacitive reactance

Substitute $1.2\text{kΩ}$ for $R$, $1.1\times {10}^3\text{Ω}$ for $X_L$, $0$ for $X_C$ to determine the impedance of the circuit,

$\begin{array}{c}Z=\sqrt{{\left(1.2\text{kΩ}\right)}^2+{\left(1.1\times {10}^3\text{Ω}-0\right)}^2}\\ =\sqrt{{\left(1.2\times {10}^3\text{Ω}\right)}^2+{\left(1.1\times {10}^3\text{Ω}\right)}^2}\\ =1.6\times {10}^3\text{Ω}\end{array}$

Conclusion:

From section 2 the impedance of the circuit is $1.6\times {10}^3\text{Ω}$.

The maximum current in the circuit is given by,

$I_m=\frac{V_m}{Z}$

• $V_m$ is the maximum voltage supplied by the source
• $Z$ is the impedance of the circuit

Substitute $1.6\times {10}^3\text{Ω}$ for $Z$, $170\text{V}$ for $V_m$ to determine the maximum current in the circuit,

$\begin{array}{c}{I}_m=\frac{170\text{V}}{1.6\times {10}^3\text{Ω}}\\ =0.11\text{A}\end{array}$

The maximum current in the circuit is $0.11\text{A}$.

To determine

The maximum voltage across the resistor and inductor.

Answer to Problem 24P

The maximum voltage across the inductor is $1.2\times {10}^2\text{V}$.

Explanation of Solution

Given Info: The resistance of the resistor is $1.2\text{kΩ}$. The inductance of the inductor is $2.8\text{H}$. The frequency of the ac circuit is $60\text{Hz}$. Maximum voltage of the source is $170\text{V}$.

The maximum voltage across the resistor is,

$\Delta V_{L,\mathrm{m}}=I_{\mathrm{m}}{X}_L$

• $I_{\mathrm{m}}$ is the maximum current in the circuit
• $X_L$ is the inductive reactance

From section (a) the inductive reactance is $1.1\times {10}^3\text{Ω}$.

Substitute $1.1\times {10}^3\text{Ω}$ for $X_L$, $0.11\text{A}$ for $I_{\text{m}}$ to determine the maximum voltage across the resistor,

$\begin{array}{c}\Delta V_{L,\mathrm{m}}=\left(0.11\text{A}\right)\left(1.1\times {10}^3\text{Ω}\right)\\ =1.2\times {10}^2\text{V}\end{array}$

Conclusion: The maximum voltage across the inductor is $1.2\times {10}^2\text{V}$.

To determine

The potential difference across the resistor, inductor, and the ac source when the current is at its maximum.

Answer to Problem 24P

The voltage across the resistor when the current is maximum is $1.3\times {10}^2\text{V}$.

Explanation of Solution

Given Info: The resistance of the resistor is $1.2\text{kΩ}$. The inductance of the inductor is $2.8\text{H}$. The frequency of the ac circuit is $60\text{Hz}$. Maximum voltage of the source is $170\text{V}$.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum voltage across the resistor is,

$\Delta V_{R,\mathrm{m}}=I_{\mathrm{m}}R$

• $I_{\mathrm{m}}$ is the maximum current in the circuit
• $R$ is the resistance

Substitute $1.2\text{kΩ}$ for $R$, $0.11\text{A}$ for $I_{\text{m}}$ to determine the maximum voltage across the resistor,

$\begin{array}{c}\Delta V_{R,\mathrm{m}}=\left(0.11\text{A}\right)\left(1.2\text{kΩ}\right)\\ =1.3\times {10}^2\text{V}\end{array}$

The voltage across the resistor when the current is maximum is $1.3\times {10}^2\text{V}$.

The current and the voltage across an inductor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum.

Hence when the current reaches maximum, the voltage across the inductor will be zero.

The voltage across the inductor when the current is maximum is $0\text{V}$.

When Kirchoff’s law is applied to the closed circuit,

$\Delta V_s-\Delta V_R-\Delta V_L=0$

• $\Delta V_s$ is the voltage across the AC source
• $\Delta V_R$ is the voltage across the resistor
• $\Delta V_L$ is the voltage across the inductor

The voltage across the AC source will be,

$\Delta V_s=\Delta V_R+\Delta V_L$

Substitute $1.3\times {10}^2\text{V}$ for $\Delta V_R$, $0$ for $\Delta V_L$ to determine the voltage across the AC source,

$\begin{array}{c}\Delta V_s=1.3\times {10}^2\text{V}+0\\ =1.3\times {10}^2\text{V}\end{array}$

Conclusion:

When the current is maximum the potential difference across the resistor, inductor, and the ac source is $1.3\times {10}^2\text{V}$, $0\text{V}$, and $1.3\times {10}^2\text{V}$ respectively.

To determine

The potential difference across the resistor, inductor, and the ac source when the current is at it is zero.

Answer to Problem 24P

When the current is zero the potential difference across the resistor, inductor, and the ac source is $0\text{V}$, $1.3\times {10}^2\text{V}$ and $1.3\times {10}^2\text{V}$ respectively.

Explanation of Solution

Given Info: The resistance of the resistor is $1.2\text{kΩ}$. The inductance of the inductor is $2.8\text{H}$. The frequency of the ac circuit is $60\text{Hz}$. Maximum voltage of the source is $170\text{V}$.

When Kirchoff’s law is applied to the closed circuit,

$\Delta V_s-\Delta V_R-\Delta V_L=0$

• $\Delta V_s$ is the voltage across the AC source
• $\Delta V_R$ is the voltage across the resistor
• $\Delta V_L$ is the voltage across the inductor

The voltage across the AC source will be,

$\Delta V_s=\Delta V_R+\Delta V_L$

Substitute $0$ for $\Delta V_R$, $1.2\times {10}^2\text{V}$ for $\Delta V_L$ to determine the voltage across the AC source,

$\begin{array}{c}\Delta V_s=0+1.2\times {10}^2\text{V}\\ =1.2\times {10}^2\text{V}\end{array}$

The current and the voltage across a resistor are in phase with each other. So when the current is minimum, the voltage across the resistor will be minimum. So when the current is zero the voltage across the resistor will be zero.

The voltage across the resistor when the current is zero is $0$ .

The current and the voltage across an inductor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum voltage across the resistor is,

$\Delta V_{L,\mathrm{m}}=I_{\mathrm{m}}{X}_L$

• $I_{\mathrm{m}}$ is the maximum current in the circuit
• $X_L$ is the inductive reactance

From section (a) the inductive reactance is $1.1\times {10}^3\text{Ω}$.

Substitute $1.1\times {10}^3\text{Ω}$ for $X_L$, $0.11\text{A}$ for $I_{\text{m}}$ to determine the maximum voltage across the resistor,

$\begin{array}{c}\Delta V_{L,\mathrm{m}}=\left(0.11\text{A}\right)\left(1.1\times {10}^3\text{Ω}\right)\\ =1.2\times {10}^2\text{V}\end{array}$

The voltage across the inductor when the current is zero is $1.2\times {10}^2\text{V}$.

When Kirchoff’s law is applied to the closed circuit,

$\Delta V_s-\Delta V_R-\Delta V_L=0$

• $\Delta V_s$ is the voltage across the AC source
• $\Delta V_R$ is the voltage across the resistor
• $\Delta V_L$ is the voltage across the inductor

The voltage across the AC source will be,

$\Delta V_s=\Delta V_R+\Delta V_L$

Substitute $0$ for $\Delta V_R$, $1.2\times {10}^2\text{V}$ for $\Delta V_L$ to determine the voltage across the AC source,

$\begin{array}{c}\Delta V_s=0+1.2\times {10}^2\text{V}\\ =1.2\times {10}^2\text{V}\end{array}$

The voltage across the AC source when the current is zero is $1.2\times {10}^2\text{V}$.

Conclusion:

When the current is zero the potential difference across the resistor, inductor, and the ac source is $0\text{V}$, $1.2\times {10}^2\text{V}$ and $1.2\times {10}^2\text{V}$ respectively.