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stability question
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<blockquote data-quote="ljb211087" data-source="post: 1272828" data-attributes="member: 564611"><p>P=VI</p><p></p><p>V=IR</p><p></p><p>P=I^2*R</p><p></p><p>When voltage is doubled in the case of bridging, the following happens</p><p></p><p>P=2VI</p><p></p><p>Therefore P=2V^2*R</p><p></p><p>For example if voltage was 10v in stereo and 20v bridged, then @ 4ohms</p><p></p><p>In bridged- (20)^2*4= 1600 watts</p><p></p><p>In stereo- (10)^2*4= 400 watts</p><p></p><p>So bridging has the effect of quadrupling the power output for a given load. As most amps cannot do this, by doubling the impedence, the power is halved which makes the amp much happier. This is why most amps can only handle twice the rated stereo impedance. Eg. 2 ohm stable per channel, 4 ohm stable when bridged</p></blockquote><p></p>
[QUOTE="ljb211087, post: 1272828, member: 564611"] P=VI V=IR P=I^2*R When voltage is doubled in the case of bridging, the following happens P=2VI Therefore P=2V^2*R For example if voltage was 10v in stereo and 20v bridged, then @ 4ohms In bridged- (20)^2*4= 1600 watts In stereo- (10)^2*4= 400 watts So bridging has the effect of quadrupling the power output for a given load. As most amps cannot do this, by doubling the impedence, the power is halved which makes the amp much happier. This is why most amps can only handle twice the rated stereo impedance. Eg. 2 ohm stable per channel, 4 ohm stable when bridged [/QUOTE]
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