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3 parts
a

6.0 V\text{6.0 V}

b

12 Ω\text{12 } \Omega

c

1.0 Ω\text{1.0 } \Omega

Step-by-step solution

8 steps

Here's the solution to the circuit problem:

Part a· 2 steps

Calculate the potential difference across the 4.0 Ω\Omega resistor.

  1. Identify the given values for the 4.0 Ω\Omega resistor.

    The current flowing through the 4.0 Ω\Omega resistor is I=1.5 AI = 1.5 \, \text{A}.
    The resistance is R=4.0 ΩR = 4.0 \, \Omega.

  2. Apply Ohm's Law (V=IRV = IR) to find the potential difference.

    V=I×RV = I \times R V=1.5 A×4.0 ΩV = 1.5 \, \text{A} \times 4.0 \, \Omega V=6.0 VV = 6.0 \, \text{V}

    The potential difference across the 4.0 Ω\Omega resistor is 6.0 V\boxed{\text{6.0 V}}.

Part b

12 Ω\text{12 } \Omega

Calculate the resistance of resistor R2.

  1. Determine the potential difference across R2.

    Resistors in a parallel branch have the same potential difference across them. Since R2 is in parallel with the 4.0 Ω\Omega resistor, the potential difference across R2 is the same as calculated in part (a).

    VR2=6.0 VV_{R2} = 6.0 \, \text{V}

  2. Determine the current flowing through R2.

    The total current entering the parallel combination is 2.0 A2.0 \, \text{A} (as shown by the current flowing through R1 and into the junction).
    By Kirchhoff's Current Law, the total current entering a junction equals the sum of currents leaving it.

    Itotal parallel=I4Ω+IR2I_{\text{total parallel}} = I_{4\Omega} + I_{R2} 2.0 A=1.5 A+IR22.0 \, \text{A} = 1.5 \, \text{A} + I_{R2} IR2=2.0 A−1.5 AI_{R2} = 2.0 \, \text{A} - 1.5 \, \text{A} IR2=0.5 AI_{R2} = 0.5 \, \text{A}

  3. Apply Ohm's Law (R=V/IR = V/I) to find the resistance of R2.

    R2=VR2IR2R_2 = \frac{V_{R2}}{I_{R2}} R2=6.0 V0.5 AR_2 = \frac{6.0 \, \text{V}}{0.5 \, \text{A}} R2=12 ΩR_2 = 12 \, \Omega

    The resistance of resistor R2 is 12 Ω\boxed{\text{12 } \Omega}.

Part c

1.0 Ω\text{1.0 } \Omega

Calculate the resistance of resistor R1.

  1. Determine the potential difference across the 2.0 Ω\Omega resistor on the right.

    The current flowing through this resistor is 2.0 A2.0 \, \text{A}.

    V2Ω=I2Ω×R2ΩV_{2\Omega} = I_{2\Omega} \times R_{2\Omega} V2Ω=2.0 A×2.0 ΩV_{2\Omega} = 2.0 \, \text{A} \times 2.0 \, \Omega V2Ω=4.0 VV_{2\Omega} = 4.0 \, \text{V}

  2. Determine the potential difference across R1 using Kirchhoff's Voltage Law.

    The total supply voltage is 12 V12 \, \text{V}. The components R1, the parallel combination, and the 2.0 Ω\Omega resistor on the right are in series.

    Vtotal=VR1+Vparallel+V2ΩV_{\text{total}} = V_{R1} + V_{\text{parallel}} + V_{2\Omega}

    We know Vtotal=12 VV_{\text{total}} = 12 \, \text{V}, Vparallel=6.0 VV_{\text{parallel}} = 6.0 \, \text{V} (from part a), and V2Ω=4.0 VV_{2\Omega} = 4.0 \, \text{V}.

    12 V=VR1+6.0 V+4.0 V12 \, \text{V} = V_{R1} + 6.0 \, \text{V} + 4.0 \, \text{V} 12 V=VR1+10.0 V12 \, \text{V} = V_{R1} + 10.0 \, \text{V} VR1=12 V−10.0 VV_{R1} = 12 \, \text{V} - 10.0 \, \text{V} VR1=2.0 VV_{R1} = 2.0 \, \text{V}

  3. Apply Ohm's Law (R=V/IR = V/I) to find the resistance of R1.

    The current flowing through R1 is 2.0 A2.0 \, \text{A}.

    R1=VR1IR1R_1 = \frac{V_{R1}}{I_{R1}} R1=2.0 V2.0 AR_1 = \frac{2.0 \, \text{V}}{2.0 \, \text{A}} R1=1.0 ΩR_1 = 1.0 \, \Omega

    The resistance of resistor R1 is 1.0 Ω\boxed{\text{1.0 } \Omega}.

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