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[ÇóÖú] Çë½ÏÈçºÎ¸ù¾ÝµçÈÝÆ÷µÄ³ä·ÅµçÇúÏß¼ÆËãinternal resistance (IR) drop

ÇëÎʱ¾°æ³¬¼¶µçÈÝÆ÷µÄ¸÷λ¸ßÈË£¬ÈçºÎ¸ù¾ÝµçÈÝÆ÷µÄ³ä·ÅµçÇúÏßÀ´¼ÆËãinternal resistance (IR) drop£¿±¾ÈËÒÔǰ×ö﮵çµÄ£¬ÏÖÔÚ¸Õ¸Õ½Ó´¥µçÈÝÆ÷·½ÃæµÄ֪ʶ£¬¶ÔÕâÒ»¼ÆËã×ÅʵºÜÃÔ㣬¸ã²»Çå£¬ÌØµØÀ´±¾°æÇë½Ì¡£
1. Çë½ÌÈçºÎ¸ù¾ÝµçÈÝÆ÷µÄ³ä·ÅµçÇúÏß¼ÆËãinternal resistance (IR) drop?
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The internal resistance was determined from the voltage drop at the beginning of each discharge:
R=¡÷V(iRΪVµÄϱê)/2i£¬
¡÷V(iRΪVµÄϱê) is the voltage drop between the first two points from its top cut-off.
This voltage drop is also referred to as the iR drop. i is the current applied.

2. ÏëÇë½Ï¸ßÈ˼òµ¥½âÊÍ»òÕß½éÉÜһϹ¦ÂʺÍÄÜÁ¿ÃܶȼÆË㹫ʽµÄÓÉÀ´¡£
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P=V*V/[4RM]   &  E=0.5CV*V/M
where V is the applied voltage ,R is the internal resistance, M is the total mass of active electrode materials used in electrochemical capactiors, and C is the measrued total capacitance of electrochemical capactiors
ÕâÓë³£¼ûµÄPºÍE¼ÆË㹫ʽÓгöÈë°¡
E=It(Ua)/m   P=I(Ua)/m
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[ Last edited by 1101030303 on 2012-1-31 at 22:49 ]
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3Â¥: Originally posted by 1101030303 at 2012-02-01 09:23:53:
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3Â¥: Originally posted by 1101030303 at 2012-02-01 09:23:53:
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7Â¥: Originally posted by myIDname at 2012-02-01 11:05:36:
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C = i /-[¦¤V/¦¤t] = i /-(slope)
where i is the current applied, [¦¤V/¦¤t] is the slope of the discharge curve after the initial iR drop.
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R = ¦¤V£¨iR£© /2i
¦¤V£¨iR£©  is the voltage drop between the first two points from its top cut-off.  This voltage drop is also
referred to as the iR drop.  i is the current applied.  
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9Â¥: Originally posted by 1101030303 at 2012-02-01 12:21:12:
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C = i /-[¦¤V/¦¤t] = i /-(slope)
where i is the current applied, [¦¤V/¦¤t] is the slope of the discharge curve after the initial iR drop.
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