Anonymous
Anonymous asked in Science & MathematicsPhysics · 10 months ago

Need help with physics problem?

A 75.0 ohm and 45 ohm resistor are connected in parallel and across a battery. The current is 0.294A. When the 45.0 ohm resistor is disconnected, the current from the battery drops to 0.116A. Determine the emf and the internal resistance of the battery.

I had thought you use the equation: E= V+ Ir but we have to determine r and the voltage we don't have. Im confused and if you can explain i'd appreciate it

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  • 10 months ago
    Favorite Answer

    This will be a simultaneous equation. You have two values so can solve for two unknowns.

    You must set up equations with these unknowns.

    OK what you are to solve for is V of the battery and R of the battery.

    When the resistors are in parallel the effective resistance is 1/ ( 1/45 + 1/75)

    V/ ( R + 1/(1/45 + 1/75) ) = 0.294

    With only the single resistor you have

    V/ (R+75) = 0.116

    From this second one you can rearrange

    V = 0.116(R + 75)

    If you substitute this for V in the first equation it becomes

    0.116(R+75) / (R + 1/(1/45 + 1/75) ) = 0.294

    1/( 1/45+1/75) = 28.125

    so substituting

    0.116 (R+75) /(R + 28.125 ) = 0.294

    Multiply both sides by ( R + 28.125)

    0.116R + 75*0.116 = 0.294 * R + 0.294* 28.125

    rearrange to get

    75*0.116 - 0.294 * 28.125 = ( 0.294-0.116)R

    divide both sides by ( 0.294-0.116)

    R = ( 75* 0.116- 0.294* 28.125) / ( 0.294- 0.116) = 2.42 Ohms

    Now substitute this back into V = 0.116( R +75)

    = 0.116*(2.42 + 75) = 8.98 V

    Check by putting both answers into the first equation.

    8.98/(2.42+ 28.125) = 2.939925 ie it is very close to 2.94 The difference comes from the simplifying the results to three significant figures rather than keeping everything to limitless precision.

  • Vaman
    Lv 7
    10 months ago

    I get V=13.78V and R internal resistance=45.865 Ohms. Check the correct values and pose the question.

  • 10 months ago

    75||45 = 28.125Ω

    We can write E = 0.294*(R+28.125)

    and E = 0.116*(R+75)

    Equate the equations for E: 0.294*(R+28.125) = 0.116*(R+75)

    Re-arrange: R(0.294-0.116) = 75*0.116 - 28.125*0.294 = 0.43125

    R = 0.43125/0.175 = 2.4228Ω

    E = 0.294(2.4228+28.125) = 8.98V

    Check

    0.116(2.4228+75) = 8.98V checks

    E ≈ 8.98V, <<<<<

    R ≈ 2.42Ω <<<<<

  • oubaas
    Lv 7
    10 months ago

    Re = R1//R2 = 75*45/(75+45) = 3375/120 = 28,125 ohm

    E /(r+Re) = 0.294

    E = 0.294r+0.294*28.125 = 0.294r+8.269

    E/(r+75) = 0.116

    E = 0.116r+75*0.116 = 0.116r+8.700

    E = 0.294r+8.269

    E = 0.116r+8.700

    0.294r+8.269 = 0.116r+8.700

    0.178r = 0.431

    r = 0.431/0.178 = 2.4228 ohm

    E = 0.116*2.4228+8.7 = 8.9810 V

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  • 10 months ago

    The equivalent resistance of the parallel combination is 1/(1/75 + 1/45) = 75*45/120

    = 28.125 ohms, so the voltage drop across this resistor combination would be

    (0.294A)(28.125 ohm) = 8.269 V.

    When only the 45-ohm resistor is connected, the voltage drop across that resistor is 5.22 V.

    Let R be the internal resistance of the battery.

    It is necessary that

    5.22V + 0.116A*R = 8.269V + 0.294A*R,

    but this makes no sense, as it gives a negative number (around -11 ohms) for the internal resistance. No wonder you're confused, the problem itself describes a physically improbable situation.

    • sojsail
      Lv 7
      10 months agoReport

      az_lender, you got it wrong. It was the 45 ohm resistor that was disconnected.

  • Alan
    Lv 7
    10 months ago

    You have E and R in series for power supply.

    Total Resistance = R_power_supply + (1/75 + 1/45)^(-1) = R_Power_supply + ((45 + 75)/ (45*75) )^(-1)

    Total Resistance = R_power_supply + 28.125

    E=IR = 0.294*(R_Power_Supply + 28.125) =0.294*R_Power_Supply +28.125*0.294

    E = 0.294*R_Power_Supply + 8.26875

    2nd situation

    E = (0.116A) (R_Power_Supply + 75)

    so 1st set two equations for E equal to find the R_Power_Supply

    0.294*R_Power_Supply +8.26875 = 0.116*R_Power_Supply + 75*0.116

    (0.294-0.116)R_Power_supply = 75*0.116 -8.26875 = 8.7 -8.26875 =

    0.178*R_Power_Supply =0.43125

    R_Power_Supply = 0.43125/ 0.178 = 2.422752809 Ohms

    E or V power supply = 0.294( 2.422752809 +28.125) = 8.981039326 Volts

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