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# In the circuit shown in Figure 4, a battery supplies a constant voltage 40 V, the inductance is 2 H, the resistance is $10 \Omega,$ and $I(0) = 0.$(a) Find $I(t).$(b) Find the current after 0.1 seconds.

## a)$$I(t)=4-4 e^{-s t}$$ b) $$\approx 1.57 \mathrm{A}$$

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Differential Equations

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### Video Transcript

Okay, We're given a circuit with a battery E with 40 volts. Constant and resistance are with 10 warms and in dr with induct INTs to Henry and they're connected in cities as condition in the diagram on Also, it is given that current at time T equals to zero is zero, and we need to calculate the current with respect toe time. So how current is wearing with respect to time and also we need to find the value of current at time. It was 2.1 2nd. Okay, so first step will be to change the circuit in tow, our equation. So we'll apply kits jobs lawyer. And it says a total voltage drop across a circuit is equals to the voltage supplied. So therefore, we'll have easy equals, two wall taste supplied, and the voltage drop across register will be ir and voltage. Drop across the doctor will be L D I by DT. Okay, so if you write this as l d a B d t plus r, i is equals toe e, we see it is a linear differential equation. Off form d Y by D X plus B y equals toa que so we can use Ah method to solve such kind of difference Allocation by finding integrating factor as Syria's to integration off P. D X and then solution is white times. Integrating factor is equals to integration off a few times integrating factor the X plus c Okay, so we'll do that toe have that solution we need to divide by el so we'll have the i d t plus our by l high is equals to e v i l Now here we have p as our by l and cue as email. Okay, so are integrating factor will be It is to integration off our by l deity and our by Ellis Constant with respect to time. So it will come out off integration and integration of oneness piece. So this is an integrating factor. Therefore a solution will be white times higher. But here are dependent variables eyes. So this is our times I f is equals two Q which is B by L times Integrating factor T d t. Okay, so evil is constant and it is to other integration Will be It is toe r l t upon our l plus c and this can be canceled. Okay, So our solution is i times integrating factor is equals toe e by our integrating factor. Now it is integrated, so we don't have to call the titles integrating factor plus c So now we need value off. See, by condition given to us What condition is time t zero Our current is zero. Okay, so therefore zero times this'll value our ears to r l are by all time zero is e by our It is to our by l zero plus c This is zero. This is one plus c So we have CS minus ive ive Uh okay, so now we can substitute this value on our solution. So we have I times here is to are by lt is e by our it is to our by l the minus e by now. If you divide by since we need value off, I we can divide by it is toe l by t and this will give us I is e by our this will become one minus e by our times it is two minus are held by t. Right, since we have positive and we divide to commit negative. Okay, So we can take it by our common, and we'll have one minus. It is two minus. R L t. This is a I. Yeah. Okay, so now we need to calculate value off. I at equals 2.1 2nd with R equals to 10. L equals two and equals to 14. So, therefore are I will be e which is 40 or are which is 10 1 minus it is two minus are just 10 or l two times 20.1. Okay, so this is four times one minus. It is two minus point five. On calculating this value, you get 1.57 MPs. So current at time D equals 2.1, so that's a solution.

Mumbai university

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Differential Equations

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