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AP Physics?

Two resistors A and B are connected in parallel across a 12V battery. The current through B is found to be 1.1A. When the two resistors 

are connected in series to the 12 V battery,

a voltmeter connected across resistor A measures a voltage of 3.2V.

Find the resistance of B.

Answer in units of Ω.

Find the resistance of A.

Answer in units of Ω.

7 Answers

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  • 1 month ago

    12v/1.1A=10.9Ω.

    12v-3.2v=8.8v.

    I=8.8/10.9Ω=.807A

    R=E/I=3.2/.807=3.96Ω

    RA & RB=3.96Ω & 10.9Ω.

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  • oubaas
    Lv 7
    1 month ago

    in parallel

    Rb = 12/1.1 ohm

    in series

    Va = 3.2 V

    Vb = 12-Va = 8.8 V

    Ra = Rb*Va/Vb = 12/1.1*3.2/8.8 = 4.0 ohm

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  • 1 month ago

    Ra = 12V/1.1A = 120/11 Ω

    When Ra and Rb are in series across 12V and Ra has 3.2V across it, Rb must have 12-3.2 = 8.8V across it. The current through both Ra and Rb: i = 8.8/(120/11) = 242/300 or about 0.807A. Therefore Ra = 3.2V/0.807A = 960/242 Ω or about 3.97Ω

    Rb = 3.97Ω or about 4.0Ω <-----

    Ra = 120/11 or about 10.9Ω <-----

    Check 12/(3.97+10.9) = 0.807A Checks

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  • 1 month ago

    Rb = Vb/Ib = 12/1.1 = 10.9 ohms. ANS.

    V = Va + Vb = 12 = 3.2 + Vb; so that Vb = 8.8 v = IRb so that I = Vb/Rb and Va = IRa so Ra = Va/I = 3.2/(Vb/Rb) = 3.2/(8.8/10.9) = 3.96 ohms. ANS

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  • 1 month ago

    I typed in the answer, but Yahoo had problems. 

    rather than retype it all, look at the screen shot below, which I managed to salvage.

    Attachment image
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  • 1 month ago

    for parallel combination i1R1= i2R2. is., i2= 1.1xR1/R2

    for series combination R1= 3.2/I and 8.8 /I, so that R1/R2= 32/88

    then solving these two eqns we can get R1=5.87 ohm and R2 = 2.13 ohm.

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  • Anonymous
    1 month ago

    V = I * R.  resistance of B = 12 V / 1.1 amp

     

    When you connect them in series, the total voltage drop across them combined is still 12V.  Since the voltage drop across A is 3.2 V, the voltage drop b is 12V - 3.2V = 8.8 V.  You know the resistance from above so you can calculate the current flow or 0.807 amps.  This is also the current flow through A since they are in series.  V = I * R, R = 3.97 ohms.

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