Question : When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature. Solution : Given Io = 0.3 * 10-6A and VF = 0.15v The current flowing through the PN diode under forward bias is I = Io (e^(40VF) -1) for germanium diode = 0.3 * 10^(-6) (e^(40*0.15) -1) = 120.73mA. For any further queries please comment below.
Superposition Theorem Requirement: Power System is a complex network having number of voltage and current sources and number of loads. Now to analyze the network we need to find voltage/current at particular branch or load. In such cases calculations for thevenin's and norton's theorem becomes difficult. We can use superposition theorem to find voltage/current at any branch or load in complex network. In this theorem only one source considered at a time and total voltage/current is sum of voltage/current obtained by considering individual source. Statement of Superposition Theorem:
Question : An LVDT is connected to an amplifier of gain 250 the output terminals of which are connected to a 5 V voltmeter. The meter's scale can be read to 1/5th of a division and has 100 divisions in all. Given that a core displacement of 0.5 mm provides an output of 2 mV from the LVDT, find, 1. The sensitivity of the LVDT 2. The sensitivity of the whole system 3. The resolution of the instrument in centimeters Explanation : (a) Sensitivity of LVDT = Output voltage /displacement Sensitivity of LVDT = 2mV/0.5 mm = 4 mV/mm (b)Sensitivity of whole setup = amplification factor*sensitivity of LVDT Sensitivity of whole setup = 250*(4 mV/mm) = 1000 mV/mm (c) The resolution of the instrument in centimeters = minimum voltage that can be read by instrument /sensitivity of whole setup There are 100 divisions to measure upto 5V. So 1 division can read 5/100 = 0.05V Scale can read 1/5 of a division. So minimum voltage that can be read = 0.05/5 ...
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