 Hello everybody, this is Mr. Mirajdar Bala Sahib, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Soulapur. In this video lecture, we will discuss applications to electrical circuit part 2. Learning outcome, at the end of this session, students will be able to solve problems of RL and RC series electrical circuits with initial condition. In previous video lecture, we discussed how to form a differential equation of RL and RC circuit along with the some examples. Now, in this video session, we will illustrate more examples on RL and RC circuits. Let us pause the video for a while and write the answer to the question. Question is write the differential equation of RC series circuit. Come back. I hope you have written answer to this question. Here I will going to explain the solution. Question is write the differential equation of RC circuit. RC circuit means a circuit containing resistance and capacitance in series with electromotive force E of t. As we know that voltage drop across resistance is R into dq by dt and voltage drop across capacitance is q by c and by using Kirchhoff's voltage law, sum of voltage drop across R and c is equal to electromotive force applied to the circuit. R into dq by dt plus q by c is equal to E of t, where E of t is a voltage source called as electromotive force. So, this is the required differential equation of RC circuit. Now, let us consider the example. Example 1, an electromotive force of 200 into e raised to minus 5 t volt is connected in series with 20 ohm resistance, 0.01 parade capacitance. Find the charge and current at a time t, if there is no charge initially solution. By reading this problem, we come to know that given resistance R equal to 20 ohm, capacitance c equal to 0.01 parade and electromotive force EMF, which is denoted by E of t equal to 200 into e raised to minus 5 t volt and initially there is no charge, means q equal to 0 at t equal to 0. By knowing this problem, we come to know that it is a RC circuit, so that the differential equation of RC circuit is R into dq by dt plus q by c equal to e of t. Now, in this substitute the value of R, c and e of t. Therefore, R is 20 ohm, therefore 20 into dq by dt plus q upon c is 0.01 parade, so q upon 0.01 is equal to value of e of t is 200 into e raised to minus 5 t. Now, dividing 20 on both the side, we get dq by dt plus q upon 20 into 0.01 is equal to 200 into e raised to minus 5 t upon 20. Now, after simplification, we get dq by dt plus 5q is equal to 10 into e raised to minus 5 t, denote this equation by 1. It is the differential equation of u 1 circuit. This differential equation belongs to first order first degree linear differential equation with q as a dependent variable and t independent variable. When we compare this differential equation with general linear differential equation dy by dx plus py equal to q, we get value of capital P is equal to 5 and value of capital Q is 10 into e raised to minus 5 t and these are the either function of t or constant. Now, the solution of differential equation 1, we have to find out. First of all, we find its integrating factor i f equal to e to power integration of P dt which is equal to e to power integration of P is 5 into dt is equal to e to power 5 as it is an integration of dt is t. Therefore, integrating factor of differential equation 1 is e raised to 5 t. Now, general solution of differential equation 1 is given by q into i f is equal to integration of i f into q into dt plus k. Now, substitute to the value of integrating factor and value of capital Q, we get q into e raised to 5 t is equal to integration of e raised to 5 t into value of q is 10 into e raised to minus 5 t dt plus k. Therefore, q into e raised to 5 t is equal to now, 10 constant we can take outside into integration of e raised to 5 t into e raised to minus 5 t becomes e raised to 0 dt plus constant k. Therefore, q into e raised to 5 t is equal to 10 into integration of dt is t and plus constant k. Now, dividing e raised to 5 t on both this side, we get charge q equal to 10 t plus k upon e raised to 5 t that is q equal to 10 t into e raised to minus 5 t plus k into e raised to minus 5 t denote this equation by 2. Now, we have to find value of constant k using initial condition we are given q is equal to 0 at t equal to 0 put it in equation second. Therefore, 0 is equal to 0 plus k into e raised to 0, but e raised to 0 is 1. So, that k equal to 0 now substitute this value of k in equation 2 we get charge q equal to 10 into t into e raised to minus 5 t coulomb this is the charge at any time t. Now, to find the current i we have to differentiate this charge q with respect to 2 t we get d q by dt equal to now in the right hand side 10 as it is and the differentiation of t into e raised to minus 5 t by product rule keeping t as it is and derivative of e raised to minus 5 t is minus 5 into e raised to minus 5 t plus keeping e raised to minus 5 t as it is and derivative of t is 1 bracket close which is equal to 10 as it is. Now, taking e raised to minus 5 t is common outside in bracket we get minus 5 t plus 1 bracket close into e raised to minus 5 t, but i is equal to d q by dt. Therefore, i is equal to 10 in bracket minus 5 t plus 1 bracket close into e to power minus 5 t ampere. This is the required current at any time t flowing in the circuit. Now, consider example 2 a condenser of capacity c is charged through resistance r by steady voltage v prove that if q equal to 0 at t equal to 0 the charge is q equal to c into v in bracket 1 minus 5 t ampere. This is the minus e to power minus t upon r c bracket close solution the given circuit is r c circuit. So that it is differential equation is r into d q by dt plus q by c equal to e of t, but we are given steady voltage v as a electromotive force e of t. Therefore, r into d q by dt plus q by c equal to v dividing r on both the side we get d q by dt plus q upon r into c equal to v upon r. So, this is the differential equation of given circuit and it is the differential equation of first order and first degree belongs to linear differential equation type dy by dx plus py is equal to q. Here p means 1 upon r into c and capital q means v upon r. Now, we can find the solution of this differential equation for that first of all we find its integrating factor by formula i f equal to e to power integration of p with respect to t which is equal to e to power integration of value of p is 1 upon r into c which is equal to e to power 1 upon r into c constant as it is and integration of dt is t. Therefore, integrating factor is equal to e to power t upon r into c. Now, general solution of this differential equation is q into integrating factor is equal to integration of integrating factor into q into dt plus constant of integration k. Now, substitute the value of integrating factor and value of capital q we get q into e to power t upon r into c is equal to integration of e to power t upon r into c into value of capital q is v by r dt plus k. Therefore, q into e raise to t upon r into c is equal to now in right side v upon r is constant taken outside the integration and integration of e raise to t upon r into c is e to power t upon r into c whole divided by 1 upon r into c plus constant k. Therefore, q into e raise to t upon r into c is equal to in right side in denominator r r gets cancel and c goes to numerator. So, that v into c into e to power t upon r into c plus constant k. Now, dividing e raise to t upon r into c on both this side we get q equal to v into c into e to power t upon r into c plus k whole divided by e to power t upon r into c. Therefore, q equal to v into c plus k into e raise to minus t upon r into c denote this equation by 2. Now, we have to find value of constant k using u 1 initial condition that is q equal to 0 at t equal to 0 this q 1 put these 2 value in equation 2 we get 0 equal to v into c plus k into e raise to 0, but e raise to 0 is 1. So, that k equal to minus v into c substitute this value of k in equation 2. Therefore, charge q equal to v into c as it is plus value of k is minus v into c into e raise to minus t by r into c. Therefore, q is equal to now taking v into c common in the right hand side we get v into c in bracket 1 minus e raise to minus t upon r into c. So, this is the required charge at any time t we have to prove hence the proof to prepare this video lecture I refer this book as a references. Thank you.