 super mesh analysis of the circuit. So, to solve a circuit with the help of K v l K c l node analysis mesh analysis, these methods can be used. So, sometimes it will happen that conditions in the circuit are not matching or not up to the mark to apply these methods. In such cases we have to use super mesh analysis of the circuit. So, today we will see what are the steps to solve the problem or to find out the unknown values with the help of super mesh analysis. Learning outcome at the end of this session students will be able to apply super mesh analysis to find unknown currents. Yes. So, to study the super mesh analysis we have to consider a simple circuit or we have to take a example of a circuit in which the condition will arise. So, that it is required to apply super mesh analysis instead of mesh analysis. Yes. So, generally in mesh analysis we will apply K v l to the loops and after solving the equations you can find out the value of unknown currents. But in the example circuit sample you can see here that between these two loops there comes a branch which consists of current source. Yes. If such condition will arise it is not possible to solve the problem with the help of mesh analysis. In such cases we will go for super mesh analysis and we will find out the unknown currents. So, now we see the steps to solve the problem with the help of super mesh analysis. So, first step is find total number of mesh for a given circuit. So, we have to find out number of mesh present in a given circuit. So, in the given circuit there are total 3 loops or mesh. Loop number 1 this is loop number 2 and outer loop this is the third loop. Yes. So, from this 3 loops if you apply mesh analysis to the 2 loops in a general method you can easily find out the unknown value of current. But here a common branch consists of current source is there. So, you cannot apply mesh analysis. Second step write K v l equation for mesh. So, instead of applying mesh analysis to these 2 loops for the outer loop you have to write down K v l equation and inside super mesh we have to apply K c l. So, for these 2 loops the branch which is common will consist of current source. So, for this particular portion we have to apply K c l or we have to write K c l equation and after solving those equations you can find out mesh currents or unknown currents is it. So, we will take one problem statement and we will try to solve it. So, this is the problem statement and we have to find out loop currents. So, we assign or I will assign loop currents as current I 1, loop current I 2 and loop current I 3. You can observe this branch is common between loop 1 and 2 there is a register of 1 ohm between loop 2 and loop 3 there exist a register of 1 ohm. But between loop 1 and loop 3 there exist a current source of 1 ampere. So, you cannot apply mesh analysis directly instead of that we have to go for super mesh. So, what we are going to do is we will apply mesh analysis to the loop 2 mesh analysis to the loop 2 and we will apply a mesh analysis to combining loop 1 and loop 3. By combining loop 1 and loop 3 we will apply mesh analysis and inside the super mesh will apply K c l. So, we will solve this. So, I will just draw the portions where we are going to apply mesh analysis or I will redraw the circuit once again. So, it is good for you to understand 1 ohm, 1 ohm, register of 1 ohm, current source 1 ampere, voltage source 1 volt, loop 1, loop 2 and loop 3. What I do is while applying mesh analysis it is required to assign the terminals of voltages to the registers. So, according to loop 1 I am assigning the terminals of the voltages to the registers. According to loop 2 these are the terminals of voltages for registers and for loop 3 plus minus plus minus. Yes, what rule we are using to assign the terminals to the voltages across the registers? Current entering the resistor the terminal is positive and current leaving the terminal of register the terminal is negative. For example, current I 3 is entering to this part of the register. So, voltage terminal is positive and leaving from this terminal of the register. So, it is negative. I have assigned the terminal to the voltages. Now, apply KVL to the loop 2. I have to apply KVL to the loop 2, register of 1 ohm is common between loop 1 and 2. So, while writing the voltage across this what we have to do is we have to add or subtract the currents. If direction of currents are same we have to add it. If direction of currents are opposite then we have to subtract it. So, here direction of current I 1 and I 2 is opposite. So, we have to subtract the current and how subtraction is to be done? What method or what assumption we have to do is to which loop you are applying the KVL that loop current is greater than the remaining currents. So, voltage across 1 ohm I will write it as a minus 1 into I 2 minus I 1. For loop 2 there are 3 registers 1, 2 and 3. So, we have to add the voltages across this registers. So, first term is this minus 1 into I 2 minus I 1 minus 1 into I 2 minus 1 into I 2 and for this particular register this register is common between loop 2 and 3. So, minus 1 into I 2 minus I 3 equal to 0. Next, apply KVL to the combined loops 1 and 3. I will draw the portion of that here. This is the portion 1 ohm 1 ohm 1 ohm plus minus 1 volt and we have to show the terminals of voltages also. This register is a part of loop 1. This register is part of loop 3 and this register is also part of loop 3. So, KVL equation is 1 1 minus this 1 into I 1 minus I 2 because this register is common between loop 1 and 2. Here minus 1 into I 3 minus I 2 this register is common between loop 3 and 2 and here it is minus 1 into I 3 equal to 0. And inside super mesh you have to apply KCL. So, this portion of the circuit this portion of the circuit it is inside the super mesh. So, we have to we have to write KCL equation. If you check the direction of this current source direction of current I 1 and direction of current I 3 direction of current I 1 is similar to direction of current source and direction of current I 3 is opposite to the direction of current source. So, applying KCL inside super mesh equation is I 1 minus I 3 is equal to 1. Now, make this as a equation 1 this as a equation 2 and this as a equation 3 solving equations 1 2 and 3 you can find out values of current I 1 I 2 and I 3 and values are 7 sorry 7 by 5 amps 3 by 5 amps and 2 by 5 amps. So, while preparing this video lecture I have used circuit theory analysis and synthesis by H. Akravarti, Dhanpatraya publication 6th edition. Thank you.