Two Resistors, Physics Question?

NetherCraft 0

Two resistors of resistances R_1 and R_2, with R_2 > R_1, are connected to a voltage source with voltage V_0. When the resistors are connected in series, the current is I_s. When the resistors are connected in parallel, the current I_p from the source is equal to 10I_s.

Let r be the ratio R_1/R_2. Find r.

Round your answer to the nearest thousandth.

Any help is appreciated, Thanks!

2 Answers

  • Series: I = Is = V/(R1+R2)

    parallel, R = R1*R2/(R1+R2)

    Ip = 10*Is = V/R = V(R1+R2)/R1*R2

    10V/(R1+R2) = V(R1+R2)/R1*R2

    10/(R1+R2) = (R1+R2)/R1*R2

    10*R1*R2 = (R1+R2)(R1+R2)

    10*R1*R2 = 2R1R2 + R1² + R2²

    8R1R2 = R1² + R2²

    r = R1/R2

    R1 = rR2

    8R1R2 = R1² + R2²

    8rR2R2 = (rR2)² + R2²

    8rR2² = (rR2)² + R2²

    R2² (8r – r² – 1) = 0

    assuming R2 is not zero

    r² – 8r + 1 = 0

    quadratic equation:

    to solve ax² + by + c = 0

    x = [-b ±√(b²-4ac)] / 2a

    r = [8 ±√(64-4)] / 2

    r = [8 ± 7.7460] / 2

    r = 0.127

    needs checking

    .

  • one of the resistors is 6 ohms the other is 4 ohms

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