Task 3 a) Using the laws of logarithms and exponentials solve the voltage equations across a battery 1. x(0.2)° – 20.6 = 10 cos (÷) sin (÷) + sin ( s() cot) 2. log,(x + 12) = 2 3. e-(x²+4x-10) = e-2...


Task 3<br>a) Using the laws of logarithms and exponentials solve the voltage equations across a<br>battery<br>1. x(0.2)° – 20.6 = 10 cos (÷) sin (÷) + sin (<br>s() cot)<br>2. log,(x + 12) = 2<br>3. e-(x²+4x-10) = e-2<br>%3D<br>correct to 4 significant figures.<br>b) Simplify the following equation using positive indices only<br>33(3x+2)4(2x+0.5)<br>2(2x-5)9(2x-2)<br>c) The current follows in a charging inductor I(t) at time t seconds is given by:<br>i(t) = 1,(1 – e mA<br>Where Iş is the supply current and t= 30.<br>1. Evaluate the following<br>The current flows in the inductor up to 3 significant figures after 24 seconds<br>if the supply current I, = 75 mA<br>The time t to 3 significant figures taken for current flows in the inductor to reach 40<br>mA if the supply current I̟ remains at 75 mA.<br>i.<br>ii.<br>2. Find an equation for the energy and evaluate it when L= 10 mH.<br>

Extracted text: Task 3 a) Using the laws of logarithms and exponentials solve the voltage equations across a battery 1. x(0.2)° – 20.6 = 10 cos (÷) sin (÷) + sin ( s() cot) 2. log,(x + 12) = 2 3. e-(x²+4x-10) = e-2 %3D correct to 4 significant figures. b) Simplify the following equation using positive indices only 33(3x+2)4(2x+0.5) 2(2x-5)9(2x-2) c) The current follows in a charging inductor I(t) at time t seconds is given by: i(t) = 1,(1 – e mA Where Iş is the supply current and t= 30. 1. Evaluate the following The current flows in the inductor up to 3 significant figures after 24 seconds if the supply current I, = 75 mA The time t to 3 significant figures taken for current flows in the inductor to reach 40 mA if the supply current I̟ remains at 75 mA. i. ii. 2. Find an equation for the energy and evaluate it when L= 10 mH.

Jun 03, 2022
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