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Ohm's Law and dB

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A power amplifier increases a signal’s voltage by 42.04 db and increases the output power by 70.0 db. the input signal is -10 dbv and the input resistance is 10kohm.

    a. What is the output signal amplitude in dB?
    b. What is the output signal amplitude in volts?

    c. What is the input voltage?
    d. Use Ohm's law (V = IR) to find the input current.

    e. What is the output current in amps?
    f. What is the load resistance?


    2. Relevant equations
    Ohm's and Power law
    V = IR
    P = IV = I2R = V2/R

    Gain (Voltage and Power)
    AvdB = 20 log(Vout/Vin)
    ApdB = 10 log(Pout/Pin)

    3. The attempt at a solution
    My issue is coming from the wording of the question. To me, it comes off a little vague.

    42.04 dB to Av = 126.47 V/V
    70.0 dB to Ap = 10,000,000 W/W

    We know the ratios off the bat and we know that:

    -10 dBV = .3162 Vrms or .894 Vpp or .447 Vp (which answers c)

    Now the problem is which voltage to use, I am assuming Vp (since a and b ask for output amplitude) so in that case for d) - which makes input current = 44.7 μA. While we are at it, we can get our power through power law, making it 19.9 μW.

    So we know the input voltage so I am assuming Av = Vout/Vin:

    126.47 = Vout/.447 = 126.47(.447) = 56.55 V (answers b)

    Now, 10,000,000 = Pout/.0000199 = 10,000,000(.0000199) = 199 W

    Using Power law, I = P/V = 3.51 A (answers e)

    Using Ohm's Law, R = V/I = 16.1 Ω (answers f)

    I am hoping someone can just check my work and see if it correct. But the big thing is a as it doesn't make sense what they are asking. I am assuming they already gave our output in the question already (42.04 dB). But I am hoping someone will give some insight.
     
  2. jcsd
  3. Nov 14, 2014 #2

    CWatters

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    Science Advisor
    Homework Helper

    I'm a bit rusty but ....

    42.04 dB is the gain. If the input is -10 dB then the output is

    -10 + 42.04 = 32.04 dBV.

    or look at it this way...

    The input voltage is 0.316 Vrms so the output is..

    0.316 * 126.47 = 39.96 Vrms

    The reference voltage for 0dBV = 1Vrms so the output in dBV is...

    20 * Log(39.96/1) = 32.04 dBV
     
    Last edited: Nov 14, 2014
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