Math Problem Statement

Refer to Figure below, determine v(t) and i(t).

Solution

The given circuit consists of a resistor, inductor, and a sinusoidal voltage source. Let's solve for i(t)i(t) and v(t)v(t), the current through and voltage across the inductor respectively.

Step 1: Write the expression for the voltage source.

The voltage source is given as: vs(t)=5cos(10t)v_s(t) = 5 \cos(10t) This means the angular frequency ω=10rad/s\omega = 10 \, \text{rad/s} and the peak voltage is 5 V.

Step 2: Impedance of the circuit elements.

  • Resistor: The resistor has a resistance R=4ΩR = 4 \, \Omega.
  • Inductor: The impedance of the inductor ZL=jωLZ_L = j\omega L, where L=0.2HL = 0.2 \, \text{H} and ω=10rad/s\omega = 10 \, \text{rad/s}. ZL=j×10×0.2=j2ΩZ_L = j \times 10 \times 0.2 = j2 \, \Omega

Thus, the total impedance ZtotalZ_{total} of the series circuit is: Ztotal=R+ZL=4+j2ΩZ_{total} = R + Z_L = 4 + j2 \, \Omega The magnitude of the impedance is: Ztotal=42+22=16+4=20=4.47Ω|Z_{total}| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 4.47 \, \Omega The phase angle θ\theta of the impedance is: θ=tan1(24)=tan1(0.5)=26.57\theta = \tan^{-1}\left(\frac{2}{4}\right) = \tan^{-1}(0.5) = 26.57^\circ

Step 3: Find the current i(t)i(t).

Using Ohm's law in the phasor domain, we can find the phasor current: I=VsZtotal=54.4726.57=1.1226.57AI = \frac{V_s}{Z_{total}} = \frac{5}{4.47} \angle -26.57^\circ = 1.12 \angle -26.57^\circ \, \text{A}

Convert this back to the time domain: i(t)=1.12cos(10t26.57)Ai(t) = 1.12 \cos(10t - 26.57^\circ) \, \text{A}

Step 4: Find the voltage across the inductor v(t)v(t).

The voltage across the inductor can be found using vL=Ldi(t)dtv_L = L \frac{di(t)}{dt}. Alternatively, we can use the fact that the inductor voltage is out of phase with the current by 90 degrees.

The phasor voltage across the inductor is: VL=IZL=1.1226.57×j2=1.12×2(9026.57)=2.2463.43VV_L = I \cdot Z_L = 1.12 \angle -26.57^\circ \times j2 = 1.12 \times 2 \angle (90^\circ - 26.57^\circ) = 2.24 \angle 63.43^\circ \, \text{V}

Convert back to the time domain: vL(t)=2.24cos(10t+63.43)Vv_L(t) = 2.24 \cos(10t + 63.43^\circ) \, \text{V}

Summary of results:

  • Current: i(t)=1.12cos(10t26.57)Ai(t) = 1.12 \cos(10t - 26.57^\circ) \, \text{A}
  • Voltage across the inductor: v(t)=2.24cos(10t+63.43)Vv(t) = 2.24 \cos(10t + 63.43^\circ) \, \text{V}

Would you like further details or have any questions?

Further questions:

  1. How is the impedance of an inductor calculated?
  2. What is the significance of phase angles in AC circuits?
  3. How does the relationship between current and voltage differ for resistors and inductors?
  4. Why does the inductor voltage lead the current by 90 degrees?
  5. How would the results change if the resistance were larger or smaller?

Tip:

In AC circuits, resistors, capacitors, and inductors all interact differently with current and voltage. Understanding their phase relationships is key to analyzing complex circuits.

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Math Problem Analysis

Mathematical Concepts

AC Circuit Analysis
Impedance
Phasor Analysis

Formulas

Ohm's Law for AC: V = IZ
Impedance of Inductor: Z_L = jωL
Phasor form: V = Vm ∠θ
Time Domain Conversion: v(t) = Vm cos(ωt + θ)

Theorems

Ohm's Law in AC Circuits

Suitable Grade Level

College Level - Electrical Engineering