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Calculate P1, P2, and P3 enter the values in the chart. Find all the unknown values of E, I, R, and P for the parallel circuit.

Computer Science

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Step-by-step solution

8 steps

1a) Calculate P1, P2, P3 and fill in the values in the chart.

  1. Use the power formula P=V×IP = V \times I for each resistor.

    For R1:

    P1=V1×I1=3.79 V×(0.579×10−3 A)P_1 = V_1 \times I_1 = 3.79 \text{ V} \times (0.579 \times 10^{-3} \text{ A}) P1=0.00219441 W≈2.19 mWP_1 = 0.00219441 \text{ W} \approx 2.19 \text{ mW}

    For R2:

    P2=V2×I2=1.79 V×(2.21×10−3 A)P_2 = V_2 \times I_2 = 1.79 \text{ V} \times (2.21 \times 10^{-3} \text{ A}) P2=0.00395759 W≈3.96 mWP_2 = 0.00395759 \text{ W} \approx 3.96 \text{ mW}

    For R3:

    P3=V3×I3=1.21 V×(1.51×10−3 A)P_3 = V_3 \times I_3 = 1.21 \text{ V} \times (1.51 \times 10^{-3} \text{ A}) P3=0.0018271 W≈1.83 mWP_3 = 0.0018271 \text{ W} \approx 1.83 \text{ mW}

  2. Fill the calculated power values into the chart.

    Voltage (V)Current (I)Resistance (R)Power (P)
    R13.79V0.579 mA10 kΩ2.19 mW
    R21.79V2.21 mA1.2 kΩ3.96 mW
    R31.21V1.51 mA2.2 kΩ1.83 mW
    Total9V3.21 mA2.8 kΩTotal Power

    1b) Find all the unknown values of E, I, R, and P for the parallel circuit.

  3. Identify known values and apply parallel circuit rules.

    In a parallel circuit, the voltage across each component is the same as the total voltage.
    Given ETotal=9 VE_{Total} = 9 \text{ V}, so E1=E2=E3=9 VE_1 = E_2 = E_3 = 9 \text{ V}.
    Given I1=7 mA=7×10−3 AI_1 = 7 \text{ mA} = 7 \times 10^{-3} \text{ A}.
    Given R2=10 kΩ=10×103 ΩR_2 = 10 \text{ kΩ} = 10 \times 10^3 \text{ Ω}.
    Given P3=10 WP_3 = 10 \text{ W}.

  4. Calculate missing values for R1.

    R1=E1I1=9 V7×10−3 A≈1285.71 Ω≈1.29 kΩR_1 = \frac{E_1}{I_1} = \frac{9 \text{ V}}{7 \times 10^{-3} \text{ A}} \approx 1285.71 \text{ Ω} \approx 1.29 \text{ kΩ} P1=E1×I1=9 V×(7×10−3 A)=0.063 W=63.0 mWP_1 = E_1 \times I_1 = 9 \text{ V} \times (7 \times 10^{-3} \text{ A}) = 0.063 \text{ W} = 63.0 \text{ mW}

  5. Calculate missing values for R2.

    I2=E2R2=9 V10×103 Ω=0.9×10−3 A=0.900 mAI_2 = \frac{E_2}{R_2} = \frac{9 \text{ V}}{10 \times 10^3 \text{ Ω}} = 0.9 \times 10^{-3} \text{ A} = 0.900 \text{ mA} P2=E2×I2=9 V×(0.9×10−3 A)=0.0081 W=8.10 mWP_2 = E_2 \times I_2 = 9 \text{ V} \times (0.9 \times 10^{-3} \text{ A}) = 0.0081 \text{ W} = 8.10 \text{ mW}

  6. Calculate missing values for R3.

    I3=P3E3=10 W9 V≈1.111 A≈1.11 AI_3 = \frac{P_3}{E_3} = \frac{10 \text{ W}}{9 \text{ V}} \approx 1.111 \text{ A} \approx 1.11 \text{ A} R3=E3I3=9 V1.111 A≈8.10 ΩR_3 = \frac{E_3}{I_3} = \frac{9 \text{ V}}{1.111 \text{ A}} \approx 8.10 \text{ Ω}

  7. Calculate total current, total resistance, and total power.

    ITotal=I1+I2+I3=(7×10−3 A)+(0.9×10−3 A)+1.111 AI_{Total} = I_1 + I_2 + I_3 = (7 \times 10^{-3} \text{ A}) + (0.9 \times 10^{-3} \text{ A}) + 1.111 \text{ A} ITotal=0.007 A+0.0009 A+1.111 A=1.1189 A≈1.12 AI_{Total} = 0.007 \text{ A} + 0.0009 \text{ A} + 1.111 \text{ A} = 1.1189 \text{ A} \approx 1.12 \text{ A} PTotal=P1+P2+P3=0.063 W+0.0081 W+10 W=10.0711 W≈10.1 WP_{Total} = P_1 + P_2 + P_3 = 0.063 \text{ W} + 0.0081 \text{ W} + 10 \text{ W} = 10.0711 \text{ W} \approx 10.1 \text{ W} RTotal=ETotalITotal=9 V1.1189 A≈8.043 Ω≈8.04 ΩR_{Total} = \frac{E_{Total}}{I_{Total}} = \frac{9 \text{ V}}{1.1189 \text{ A}} \approx 8.043 \text{ Ω} \approx 8.04 \text{ Ω}

  8. Present the final table with all calculated values.

    Voltage (E)Current (I)Resistance (R)Power (P)R19 V7 mA1.29 kΩ63.0 mWR29 V0.900 mA10 kΩ8.10 mWR39 V1.11 A8.10 Ω10 WTotal9 V1.12 A8.04 Ω10.1 W\boxed{ \begin{array}{|c|c|c|c|c|} \hline & \text{Voltage (E)} & \text{Current (I)} & \text{Resistance (R)} & \text{Power (P)} \\ \hline \text{R1} & \text{9 V} & \text{7 mA} & \textbf{1.29 kΩ} & \textbf{63.0 mW} \\ \text{R2} & \text{9 V} & \textbf{0.900 mA} & \text{10 kΩ} & \textbf{8.10 mW} \\ \text{R3} & \text{9 V} & \textbf{1.11 A} & \textbf{8.10 Ω} & \text{10 W} \\ \text{Total} & \text{9 V} & \textbf{1.12 A} & \textbf{8.04 Ω} & \textbf{10.1 W} \\ \hline \end{array} }
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