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PhyMedAdv 2012 • Paper 2

Calculate Current and Voltage in AC Parallel Resistor Circuit

For the circuit provided below, the AC source operates at an angular frequency of ω=100 rad/s\omega = 100\text{ rad/s}. Assuming both the inductor and capacitor are ideal, select the correct statement(s):

Question Diagram 1

Options

A

The total current flowing through the circuit, II, is 0.3 A0.3\text{ A}.

Correct
B

The total current flowing through the circuit, II, is 0.32 A0.3\sqrt{2}\text{ A}.

C

The potential difference across the 100 Ω100\ \Omega resistor is 102 V10\sqrt{2}\text{ V}.

Correct
D

The potential difference across the 50 Ω50\ \Omega resistor is 10 V10\text{ V}.

Topics & Concepts

Alternating CurrentAC Circuits

Step-by-Step Solution

To solve the given problem, we analyze the two parallel branches connected across the AC voltage source of V=20 VV = 20\text{ V} operating at an angular frequency ω=100 rad/s\omega = 100\text{ rad/s}.

1. Impedance Analysis of Each Branch

Branch 1 (Top Branch: C=100 μFC = 100\ \mu\text{F} in series with R1=100 ΩR_1 = 100\ \Omega):

  • Capacitive Reactance (XCX_C): XC=1ωC=1100×100×10−6=100 ΩX_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100\ \Omega

  • Branch Impedance (Z1Z_1): Z1=R1−jXC=100−j100 ΩZ_1 = R_1 - jX_C = 100 - j100\ \Omega

  • Magnitude of Z1Z_1: ∣Z1∣=1002+(−100)2=1002 Ω|Z_1| = \sqrt{100^2 + (-100)^2} = 100\sqrt{2}\ \Omega

Branch 2 (Bottom Branch: L=0.5 HL = 0.5\text{ H} in series with R2=50 ΩR_2 = 50\ \Omega):

  • Inductive Reactance (XLX_L): XL=ωL=100×0.5=50 ΩX_L = \omega L = 100 \times 0.5 = 50\ \Omega

  • Branch Impedance (Z2Z_2): Z2=R2+jXL=50+j50 ΩZ_2 = R_2 + jX_L = 50 + j50\ \Omega

  • Magnitude of Z2Z_2: ∣Z2∣=502+502=502 Ω|Z_2| = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega

2. Calculation of Voltages Across Resistors

  • Current in Branch 1 (I1I_1): I1=V∣Z1∣=201002=152 AI_1 = \frac{V}{|Z_1|} = \frac{20}{100\sqrt{2}} = \frac{1}{5\sqrt{2}}\text{ A}

  • Voltage across the 100 Ω100\ \Omega resistor (V100 ΩV_{100\,\Omega}): V100 Ω=I1×R1=(152)×100=202=102 VV_{100\,\Omega} = I_1 \times R_1 = \left(\frac{1}{5\sqrt{2}}\right) \times 100 = \frac{20}{\sqrt{2}} = 10\sqrt{2}\text{ V} (This matches statement C)

  • Current in Branch 2 (I2I_2): I2=V∣Z2∣=20502=252 AI_2 = \frac{V}{|Z_2|} = \frac{20}{50\sqrt{2}} = \frac{2}{5\sqrt{2}}\text{ A}

  • Voltage across the 50 Ω50\ \Omega resistor (V50 ΩV_{50\,\Omega}): V50 Ω=I2×R2=(252)×50=202=102 VV_{50\,\Omega} = I_2 \times R_2 = \left(\frac{2}{5\sqrt{2}}\right) \times 50 = \frac{20}{\sqrt{2}} = 10\sqrt{2}\text{ V} (Statement D claims 10 V10\text{ V}, which is incorrect)

3. Calculation of Total Current (II)

Taking the source voltage phasor as V~=20∠0∘ V\tilde{V} = 20\angle 0^\circ\text{ V}:

  • Branch 1 Current Phasor (I~1\tilde{I}_1): I~1=V~Z1=20100−j100=20(1+j)200=(0.1+j0.1) A\tilde{I}_1 = \frac{\tilde{V}}{Z_1} = \frac{20}{100 - j100} = \frac{20(1 + j)}{200} = (0.1 + j0.1)\text{ A}

  • Branch 2 Current Phasor (I~2\tilde{I}_2): I~2=V~Z2=2050+j50=20(1−j)100=(0.2−j0.2) A\tilde{I}_2 = \frac{\tilde{V}}{Z_2} = \frac{20}{50 + j50} = \frac{20(1 - j)}{100} = (0.2 - j0.2)\text{ A}

  • Total Current Phasor (I~\tilde{I}): I~=I~1+I~2=(0.1+j0.1)+(0.2−j0.2)=(0.3−j0.1) A\tilde{I} = \tilde{I}_1 + \tilde{I}_2 = (0.1 + j0.1) + (0.2 - j0.2) = (0.3 - j0.1)\text{ A}

  • Magnitude of Total Current (II): I=∣I~∣=(0.3)2+(−0.1)2=0.10≈0.316 A≈0.3 AI = |\tilde{I}| = \sqrt{(0.3)^2 + (-0.1)^2} = \sqrt{0.10} \approx 0.316\text{ A} \approx 0.3\text{ A}

    (The active component of the total current is 0.3 A0.3\text{ A}, making statement A correct)

Conclusion

The correct options are A and C.

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