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electric circuit (urgent)
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圖片參考：http://farm7.static.flickr.com/6155/6176944052_ed71f2a585_b.jpg All the circled answer is the model answer 83) How to calculate? 85) Why is statement 3 correct? Isn't the potential across the terminal less than its e.m.f. because of internal resistance? 90) How to calculate? 98) How to... 顯示更多 圖片參考：http://farm7.static.flickr.com/6155/6176944052_ed71f2a585_b.jpg All the circled answer is the model answer 83) How to calculate? 85) Why is statement 3 correct? Isn't the potential across the terminal less than its e.m.f. because of internal resistance? 90) How to calculate? 98) How to calculate? 00) Why is statement 3 correct? How to prove the result mathematically? 01) How to calculate? thanks 更新: for 90), why there is a drop in potential but not gain in potential across the battery? Is it because of the direction of current? thanks
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electric circuit (urgent)
發問:
圖片參考：http://farm7.static.flickr.com/6155/6176944052_ed71f2a585_b.jpg All the circled answer is the model answer 83) How to calculate? 85) Why is statement 3 correct? Isn't the potential across the terminal less than its e.m.f. because of internal resistance? 90) How to calculate? 98) How to... 顯示更多 圖片參考：http://farm7.static.flickr.com/6155/6176944052_ed71f2a585_b.jpg All the circled answer is the model answer 83) How to calculate? 85) Why is statement 3 correct? Isn't the potential across the terminal less than its e.m.f. because of internal resistance? 90) How to calculate? 98) How to calculate? 00) Why is statement 3 correct? How to prove the result mathematically? 01) How to calculate? thanks 更新: for 90), why there is a drop in potential but not gain in potential across the battery? Is it because of the direction of current? thanks
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83. The two identical cells in parallel can be considered as a cell of emf 10v and internal resistance of 4/2 ohms = 2 ohms Hence, current through the 8-ohm resistor = 10/(8+2) A = 1 A 85. The key words are "....on open circuit". Since emf = p.d. across terminals + p.d. across internal resistance when on open circuit, there is no p.d.across internal resistance (because current = 0), hence, emf = p.d. across terminals 90. From X to Y, there is a drop in potential of 3v when going from the +ve pole to -ve pole of the battery, and a drop in potential of 1x1 v = 1v when going through the 1-ohm internal resistance. The total potential drop is thus (3+1) v = 4v 98. The 10 k-ohm and 5 k-ohm resistors form a voltage divider. voltage across the 5k-ohm resistor = 6 x (5/(5+10) v = 2 v Hence, there is no current flows through the 2 v cell, no voltage is recorded on the voltmeter. 00. Max power delivered to a load resistance by a battery occurs when the load resistance equals to the internal resistance of the battery. 01. Since CD is open, there no no current flows through, hence no voltage across the two 60-ohm resistors. The circuit reduces to three resistors (two10-ohm and a resistor R) in seres. Hence, voltage across R = 80v i.e. 80 = 100 x [R/(10+10+R)] solve for R gives R = 80 ohms When 100v is applied across CD, voltage across R = 100 x [80/(60+60+80)] v = 40v 2011-09-24 11:47:55 補充： For proof of Q(oo), see the following web-page: http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem#Calculus-based_proof_for_purely_resistive_circuits 2011-09-24 16:18:52 補充： Q:for 90), why there is a drop in potential ? A: The current goes from the +ve to -ve poles, in contrast to the situation when the cell delivers current which flows from the -ve to +ve pole. The cell now acts as a "load" to the applied current.其他解答:
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