Solve Ohm's & dB Homework: Find Volt, Current, Load Res.

In summary: BV is the output voltage in dBV. To convert to Vrms...10 ^ (32.04/20) * 1 = 39.96 VrmsTo convert to VpVp = 39.96 * 2 = 79.92 VpTo convert to VppVpp = 79.92 * 2 = 159.84 VppSo the answer to a is 32.04 dBVb. What is the output signal amplitude in volts?The output signal amplitude in volts is 39.96 Vrms or 79.92 Vp or 159.84 Vpp.c. What is the input voltage?
  • #1
ThatGuyYeah
3
0

Homework Statement


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?


Homework 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)

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.
 
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  • #2
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:

Related to Solve Ohm's & dB Homework: Find Volt, Current, Load Res.

1. What is Ohm's Law and how do you use it to solve for voltage, current, and load resistance?

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points. This means that as voltage increases, current also increases, and as voltage decreases, current decreases. To solve for voltage, current, and load resistance, we can use the formula V=IR, where V is voltage, I is current, and R is resistance.

2. How do you convert between decibels (dB) and voltage?

To convert between decibels and voltage, we can use the formula dB = 20log(V/V0), where V is the voltage being measured and V0 is the reference voltage. To find the voltage, we can rearrange the formula to V = V0 * 10^(dB/20).

3. What is the difference between voltage, current, and load resistance?

Voltage is the measure of electric potential difference between two points. Current is the flow of electric charge through a conductor. Load resistance is the resistance to the flow of current in a circuit, usually caused by components such as resistors, capacitors, and inductors.

4. Can Ohm's Law be used to solve for any type of circuit?

Yes, Ohm's Law can be used to solve for voltage, current, and resistance in any type of circuit. However, it is important to note that Ohm's Law assumes a linear relationship between voltage and current, which may not always be the case in more complex circuits.

5. What are some real-world applications of solving Ohm's Law and using decibels?

Solving Ohm's Law and using decibels is crucial in designing and analyzing electrical circuits, as well as troubleshooting any issues that may arise. It is also used in industries such as telecommunications, audio engineering, and electronics manufacturing.

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