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Electronics Revision Worksheet # 2 Solutions Solution 1 Solution 5 Note you needed this graph Solution 2 a) Y and Z are non ohmic, the graphs are not straight (linear) b) 80/510-3 = 16k c) Use the graph to see that when device Z has 2mA it has 50V. Therefore the device X must have 50V over it as the two devices are in parallel. If device X has 50V over it, use the graph and you see that the current must be 3.1~3.2mA. Use V=IR and you would find that the current is 3.13mA. (from question 2 R = 16k) a) (i) The switch on voltage is the potential difference that the diode requires in order to work. (ii) The switch on voltage for this diode is 0.7V (iii) Silicon (iv) If the diode is reverse biased the Vy > Vx (v) If the diode is forward biased the conventional current flows from X to Y b) There are two diodes, so each will take 0.7V. So the two diodes take 1.4V. Leaving 0.1V for the resistor. The current would be 0.1/60 = 0.001666A. Use this current and multiply by the voltage 1.5V and you will know the power. 2.5mW. Solution 3 a) Answer is E. The new current would be much less then half the old current. Note the graph for a diode is not linear. b) Answer is C, the new current would be similar to the old current, which would almost be 0A. Solution 4 a) (i) The diode is forward biased during the BC period. (ii) During the time interval A the diode is reversed biased. b) A sine graph Solution 6 Solution 7 a)This is the classic de-coupler, where the D.C. voltage goes over the capacitor and the A.C. voltage goes over the resistor b) Vout and Vin are parallel so Vout is Vin Solution 8