 Namaste, welcome to the session Solving Problems Based on Mathematical Modelling of Electrical Elements. At the end of this session students will be able to solve problems based on mathematical modeling of electrical systems. Now, in this session we are going to see importance of Laplace transform and transfer function then we are going to find out transfer function of a RLC circuit then examples on mathematical modeling of electrical systems. Now, let us see importance of Laplace transform and transfer function. The transformation technique relating the time functions to frequency dependent functions of a complex variable is called the Laplace transformation technique. It is useful in solving linear differential equations and analyzing the control systems. The transfer function gives mathematical models of all system components and overall system also. So, individual analysis of various system components is also easy and possible by the transfer function of purge. Now, let us see what do we mean by transfer function. So, transfer function is defined as the ratio of Laplace transform of output response of the system to the Laplace transform of input excitation with zero initial conditions. Now, let us consider here a system which has reference input as R of t and controlled output as C of t which are represented in time domain and when you take Laplace transform then the same input and output are represented by capital R of s and output as capital C of s. Then the transfer function T of s is given as T of s is equal to Laplace transform of output divided by Laplace transform of input which is equal to C of s divided by R of s. Now, let us see mathematical model of electrical elements. So, there are three basic components register inductor capacitor. So, symbol of register is given as shown in figure. Then voltage current relation for register is V is equal to R i, V is equal to i r then Laplace transform for register is V is equal to R into i of s. Then for inductor the voltage current relation is given as V is equal to L into di by dt then Laplace transform is given as V is equal to L into s into i of s. For capacitor the voltage current relationship is V is equal to 1 upon C integration i dt then Laplace transform for capacitor is given as V is equal to 1 upon C into s into i of s. Now, let us see transfer function of RLC circuit. Now, let us have example here obtained mathematical model of the given electrical system. So, this is the given electrical system which has RLC components in series. So, take a pause here and obtain mathematical model. So, let us consider the current i is flowing through the circuit. So, when current i flows through the circuit we get voltages across each component as V r, V c and V l then applying Kirchhoff's voltage law the equation becomes V is equal to V r plus V l plus V c. Now, writing mathematical model for this circuit which gives us V is equal to R into i plus L into di by dt plus 1 upon C integration i dt. Now taking Laplace on both sides of equation 1 will give us V of s is equal to R into i of s plus L s into i of s plus 1 upon C s into i of s. So, this is the equation where we are taking i of s common. So, equation becomes V of s is equal to i of s into R plus L s plus 1 upon C s. So, transfer function can be of output current to input voltage that is i of s divided by V of s equal to C s divided by L c s square plus R c s plus 1. Now, to have a transfer function of output voltage to input voltage the output voltage is nothing but voltage across capacitor. The output voltage across capacitor is V c of t is equal to 1 upon C integration i of t dt. Now, taking Laplace transform on both sides of equation that will give us V c of s is equal to 1 upon C s into i of s. So, the same equation can be written as i of s is equal to V c of s into C s. Now, substituting value of i s in equation 2 that is equation 2 is i of s divided by U of s is equal to C s divided by L c s square plus R c s plus 1. So, the equation becomes V c of s divided by V of s is equal to 1 upon C s into C s divided by L c s square plus R c s plus 1. So, cancelling C s will get the equation as V c of s divided by V of s equal to 1 upon L c s square plus R c s plus 1. Now, based on this let us solve some example. So, obtain mathematical model of the given electrical system. So, here is the given electrical system. So, let us consider current i 1 and i 2. So, applying Kirchhoff's voltage law we get the equation for the first loop as V is equal to R 1 into i 1 plus 1 upon C 1 i 1 minus i 2 dt. Now, for the second path we will get equation whereas, there is no voltage. So, 0 is equal to R 2 into i 2 plus L into d i 2 by dt plus 1 upon C 2 integration of i 2 dt plus 1 upon C 1 integration i 2 minus i 1 dt. Now, taking Laplace transform of above equation will give us V is equal to R 1 into i 1 of s plus 1 upon C 1 s into the bracket i 1 of s minus i 2 of s equation 3. Then taking Laplace transform of second equation that will give us 0 is equal to R 2 into i 2 of s plus L s i 2 of s plus 1 upon C 2 s into i 2 s plus 1 upon C 1 s into the bracket i 2 s minus i 1 s. Now, let us have second example find the transfer function of the following network. So, this is the network where we want to find out transfer function. So, let us consider current i of t flowing through the circuit. Then applying Kirchhoff's voltage law that will give us input voltage i of t is equal to R 1 i of t plus 1 upon C integration i of t dt plus R 2 into i of t for output voltage E 0 t which gives us E 0 t is equal to 1 upon C integration i of t into dt plus R 2 i of t. Now, taking Laplace transform of above equations will give us E i of s equal to R 1 i s plus 1 upon C s i of s plus R 2 i of s and E o of s is equal to 1 upon C s of i of s plus R 2 i of s equation 4. Now, from above equation equation 3 we are taking i of s is common. So, equation becomes i of s is equal to E i of s upon R 1 plus R 2 plus 1 upon C s equation 5. Then from the second equation then from the fourth equation we are taking i of s common then equation becomes E of s equal to 1 upon C s plus R 2 into i of s. Then substituting the value of i of s from equation 5 in equation 6 that will result into E of s is equal to 1 upon C s plus R 2 into E of s divided by R 1 plus R 2 plus 1 upon C s then the transfer function is nothing but output voltage upon input voltage that is E o of s divided by E i of s that will result into R 2 C s plus 1 divided by R 1 plus R 2 into C s plus 1. So, in this way we can find out mathematical modeling of electrical system as well as we can find out transfer function of electrical system. These are references. Thank you.