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Part 1: Equivalent Resistance in a Parallel Circuit

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43 Ω or 1.33 Ω\frac{4}{3} \, \Omega \text{ or } 1.33 \, \Omega

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Part 1· 4 steps

Equivalent Resistance in a Parallel Circuit

The formula for the equivalent resistance (ReqR_{eq}) of resistors connected in parallel is given by:

1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

From the image, the given resistance values are R1=4 ΩR_1 = 4 \, \Omega, R2=6 ΩR_2 = 6 \, \Omega, and R3=3 ΩR_3 = 3 \, \Omega.

  1. Substitute the given resistance values into the formula.

    1Req=14 Ω+16 Ω+13 Ω\frac{1}{R_{eq}} = \frac{1}{4 \, \Omega} + \frac{1}{6 \, \Omega} + \frac{1}{3 \, \Omega}

  2. Find a common denominator for the fractions.

    The least common multiple (LCM) of 4, 6, and 3 is 12.

    1Req=312 Ω+212 Ω+412 Ω\frac{1}{R_{eq}} = \frac{3}{12 \, \Omega} + \frac{2}{12 \, \Omega} + \frac{4}{12 \, \Omega}

  3. Add the fractions.

    1Req=3+2+412 Ω\frac{1}{R_{eq}} = \frac{3+2+4}{12 \, \Omega} 1Req=912 Ω\frac{1}{R_{eq}} = \frac{9}{12 \, \Omega}

  4. Simplify the fraction and take the reciprocal to find ReqR_{eq}.

    1Req=34 Ω\frac{1}{R_{eq}} = \frac{3}{4 \, \Omega} Req=43 ΩR_{eq} = \frac{4}{3} \, \Omega

    As a decimal, this is approximately:

    Req≈1.33 ΩR_{eq} \approx 1.33 \, \Omega

    The student's calculation for the sum of fractions (1372\frac{13}{72}) was incorrect, leading to an incorrect final resistance value.

    The correct equivalent resistance is 43 Ω or 1.33 Ω\boxed{\frac{4}{3} \, \Omega \text{ or } 1.33 \, \Omega}.

Part 2

Voltage in a Parallel Circuit

The image states: VT=V1=V2V_T = V_1 = V_2 and then VT=12=6V_T = 12 = 6.

In a parallel circuit, the voltage across each resistor is the same as the total voltage supplied by the source. Therefore, the statement VT=V1=V2V_T = V_1 = V_2 (and also V3V_3) is correct.

However, the statement VT=12=6V_T = 12 = 6 is contradictory. It implies that the total voltage is simultaneously 12 V and 6 V, which is not possible. Without additional context from the original problem, it is not possible to determine the correct total voltage.

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