An audio oscillator (ac source) with an internal resistance of 2000 Ohm and open-circuit output voltage
of 12 V is to be used to drive a loudspeaker with a resistance of 8 Ohm. What should be the ratio of
primary to secondary turns of a transformer, so that maximum power is transferred to the speaker?
Suppose a second identical speaker is connected in parallel with the first speaker. How much power is
then supplied to the two speakers combined?
SOLUTION
In a simple circuit, maximum power transfer from a source to a load requires that the load resistance
equal the internal resistance of the source. We can use Ohm’s law, the relationship between primary and
secondary currents, the primary and secondary voltages and the turns ratio of the transformers to derive
an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance
of the speaker(s).
Start with the primary transformer where the loudspeaker’s resistance is:
𝑉1
𝑅𝑒𝑓𝑓 =
𝐼1
But we can relate V1 and I1 to V2 and I2
𝑁1
𝑁2
𝑉1 = 𝑉2
𝑎𝑛𝑑 𝐼1 = 𝐼2
𝑁2
𝑁1
Substituting these expressions in above:
𝑁
𝑉2 𝑁1
𝑉2 𝑁1 2
2
𝑅𝑒𝑓𝑓 =
= ( )( )
𝑁
𝐼2 𝑁2
𝐼2 𝑁2
1
We can now solve for the ratio of turns:
𝐼2 𝑅𝑒𝑓𝑓
𝑅𝑒𝑓𝑓
𝑁1
2000Ω
=√
=√
=√
= 15.8
𝑁2
𝑉2
𝑅2
8Ω
The question now asks us to have a second speaker in parallel to the first speaker – and determine the
power to the two speakers combined. From the two speakers, in parallel, we’ll have an equivalent
speaker resistance of Rsp=4Ω (two 8Ω speakers in parallel).
With the same transformer (same ratio of turns) we can find the new Reff:
𝑁1 2
𝑅𝑒𝑓𝑓 = 𝑅2 ( ) = 4Ω(15.8)2 = 999Ω
𝑁2
Now, let’s look at the situation with respect to power:
𝑃𝑠𝑝 = 𝐼12 𝑅𝑒𝑓𝑓
We need to find the current drawn from the source:
𝑉1
12𝑉
𝐼1 =
=
= 12𝑚𝐴
𝑅𝑒𝑓𝑓 999Ω
So, the power supplied is 𝑃𝑠𝑝 = 𝐼12 𝑅𝑒𝑓𝑓 = (12𝑚𝐴)2 999Ω = .14𝑊