Power on resistos - brainteaser

Resistor is 8 Ohm.

You're not providing enough information to answer the question without making assumptions.

The total power in white noise depends on bandwidth: white noise contains equal power per unit bandwidth.

The Decibel is a logarithmic ratio i.e. it is relative, so we have to make assumptions about what the -40 and -36dB are relative to. Let's call it "line level".

Let's assume a "line level" white noise input produces 100W across an 8-ohm load. If we then inserted a total 76dB of attenuation, the output would fall to 2.5 micro-Watts. That would be 0.56mA (RMS) in an 8-ohm load.

P = 100 * 10^(-76/10) = 2.5e-6

P = I*I*R = I*I*8

I = 0.56e-3

what value is the resistor?

martin

Without knowledge of the crest factor of the white noise you can't even begin a calculation.

Do you *really* mean a -76dB signal btw ? Also be aware that any dB panel markings on an amplifier are unlikely to be very accurate at the -40 level.

Graham

Assume that setting the amplifier to -40 dB is done by an attenuator or pot at the input - this ensures that the amplifier is far from saturation. -40 dB wrt 100 W is 10 mW.

A white noise source produces a flat spectrum from near enough DC to the highest measurable frequency. In that case the bandwidth of the amplifier would only let a tiny fraction of the power through, so the power in the load would be negligible. It's more likely that the noise source produces flat noise over the bandwidth of the amplifier under test. In which case the CD player full output (0 dB) would produce 10 mW from the amplifier but at -36 dB would produce only 2.5 microwatts. The current is whatever is necessary to produce 2.5 microwatts in the load resistor.

(I was always taught that dBs is dBs!)