Question

6. Suppose a 6 C2 loudspeaker is driven from an amplifier under matched conditions, meaning that the amplifier's output impedance equals 6 0. The amplifier supplies an unloaded output voltage V-24 V a. Under these conditions what is the maximum power delivered to the loudspeaker 6 Y-24Y b. The user decides to connect a second speaker in parallel to the first to better distribute sound in the room. If one desires to deliver half the maximum original power equally to each speaker, what must be the impedance of the speakers? Prove it!

Answer 1

Rnt RLood Tam Thunem Maximum o FAOM Max. Pousn lo Thangeund to dha Loed (laudapaber) Kint I 632 Loudaeakn CLOAD BLoad G 24V Ampitin 4 x 6 I R 24 W Pmax maw. 2 24W A Lod (dee ciscuit obou) In makhed condition, Rin (mplitie) Rond (akon) given 24 W owe So dativerad to udeakn Rint Cut SS do ampliter Lot mpidane apsakna Be 2 SHeeakn Connadted 6 n 24v E Be R each LOAP(RL Now Zhrakvy in Nt Lead Fmpdance RiIRz RXR ミ(R RtR tegethn farallel eomes- loed as PROOF Now, I R1 P (acus loed (NetR 2 V Rint +R dfd VRL fr Max tour hamn dr d - 2 (RL+ent) , 2 (+li v (Rit RLtint)2RLR Rirt Pormax P Thanter 2 RRA Kiat > y4 2 V24x R Linttk Poveer =12W So, 2W (aciums) + 12W (atuws,)- 24W.