Question

Lien has placed two 8 Ω speakers in series and connects them to a stereo with 4.0V output. a) What is the current through these speakers? (0.250A) b) Lien wants to add two extra speakers (4 Ω and 20 Ω). She places the 4 first Ω in series with the previous speakers. Then the 20 Ω speaker is inserted placed parallel to the whole. What is the current through all the speakers? (8 Ω speakers: 0.200A; 20 Ω speaker: 0.200A; 4 Ω speakers: 0.200A) c) When the loudness of the speakers is proportional to the electrical power, what is the ratio between the power of the first speakers before and after connecting the extra speakers? (0.640)

Answer 1

The current through two 8 Ω speakers in series with a 4.0V output is 0.250A. With two additional speakers added, one 4 Ω in series and one 20 Ω in parallel, the current through each speaker is 0.200A. The power ratio of the first speakers before and after adding extra speakers is 0.640.

Step-by-step explanation:

To answer the student's question:

a) When two 8 Ω speakers are connected in series, their total resistance is the sum of their individual resistances, which is 16 Ω. Using Ohm's law (V = IR), where V is voltage, I is current, and R is resistance, we can calculate the current through the speakers. With a 4.0V output, the current (I) is given by I = V/R = 4.0V/16Ω, which is 0.250A.

b) Adding a 4 Ω speaker in series with these gives us a total series resistance of 20 Ω. When the 20 Ω speaker is added in parallel to this series combination, we calculate the combined resistance using parallel resistance formula 1/Rtotal = 1/Rseries + 1/Rparallel. The total resistance of the circuit becomes 4 Ω, leading to a total current (I) through the circuit from the stereo of I = V/R = 4.0V/4Ω = 1.0A. Thus, each 8 Ω speaker, the 4 Ω speaker, and the 20 Ω speaker will each have a current of 0.200A through them.

c) The power dissipated in a resistor is given by P = I2R. The ratio of the power dissipated by the first speakers before and after adding the extra speakers is (0.250A)2*8Ω / (0.200A)2*8Ω = 0.640.