14pt v BIU.. | X2 X | A | A $ AA). Ti 213141516171819 10 11 12 13 14 15 16 17 1 1 1. L Figure 1a below shows a semicondu

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answerhappygod
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14pt v BIU.. | X2 X | A | A $ AA). Ti 213141516171819 10 11 12 13 14 15 16 17 1 1 1. L Figure 1a below shows a semicondu

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14pt V Biu X2 X A A Aa Ti 213141516171819 10 11 12 13 14 15 16 17 1 1 1 L Figure 1a Below Shows A Semicondu 1
14pt V Biu X2 X A A Aa Ti 213141516171819 10 11 12 13 14 15 16 17 1 1 1 L Figure 1a Below Shows A Semicondu 1 (69.27 KiB) Viewed 15 times
14pt v BIU.. | X2 X | A | A $ AA). Ti 213141516171819 10 11 12 13 14 15 16 17 1 1 1. L Figure 1a below shows a semiconductor diode. Use this Figure to answer question 1a to 1c below. А 01 P N D2 6V 5K Figure 1a: For question 1a - 1c Figure 1b: For question 1d (a) If point B and C are connected to the positive and negative terminal of a battery, then state the correct answers to the following under this condition: 0 What is this biasing condition called? (5 Marks) (1) What happens to the size of region A? (5 Marks) What happens to the resistance of region A? (10 Marks) (I)
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