 going to see frequency for maximum voltage across capacitor in case of series resonance. So, in this particular derivation we are going to find out value of frequency at which voltage across capacitor is maximum. Learning outcome at the end of this session students will be able to demonstrate how to find out frequency at which voltage across capacitor is maximum in case of series resonance. Before starting with this friends you should recall what is the relation between frequency and reactances as well as you should know what is the formula for resonant frequency. Now, we will start with the derivation for frequency at which voltage across capacitor is maximum. So, while finding frequency for maximum voltage across capacitor we should first write down the equation for v c voltage across capacitor. So, voltage is represented by current multiplied by capacitive reactance is it and capacitive reactance is represented by 1 upon omega into c is it as well as current is represented by voltage upon impedance voltage upon impedance. So, v c is represented by v upon omega c I will directly write down the magnitude of impedance that is r square plus omega l minus 1 upon omega c bracket square is it. So, if we take square on both side. So, v c square is equal to v square upon omega square c square and in bracket it is r square plus omega square l square c omega square l c minus 1 whole bracket square upon omega square c square is it. I have solved this bracket next step is v c square is equal to v square. Now, if we multiply this omega square c square to this particular bracket it will become omega square c square r square plus omega square l c minus 1 bracket square is it. Now, differentiating v c square with respect to omega and equating numerator term to 0 and equating numerator term to 0 means what we have to do is d v c square by d omega equal to 0 this we have to do. Now, v c square is equal to v square upon omega square c square r square plus omega square l c minus 1 bracket square. It is in the form that is u by v and if it is in the form u by v we can write it or we can solve the difference or we take the differentiation as v d u minus u d v v d u minus u d v v d u if you take the differentiation of constant it become 0 minus u d v term. I will directly write down the u d v term and it is v square in bracket 2 omega c square r square plus 2 in bracket omega square l c minus 1 into 2 omega l c and that we have to equate to 0 is it. We further simplify it as 2 omega c square r square plus 4 omega l c in bracket omega square l c minus 1 equal to 0 voltage will not be 0. So, next term is 2 omega c square r square plus 4 omega cube l square c square minus 4 omega c square r square plus 2 omega c square r square plus omega l c equal to 0 is it. Now, I take these two terms to the RHS except this. So, 4 omega cube l square c square is equal to 4 omega l c minus 2 omega c square r square. Now, I keep omega square from these terms common that is 4 omega cube l square c square from this I take omega cube term common and remaining term I will take to the RHS. So, it is 4 omega l c upon 4 omega l square c square minus 2 omega c square r square upon 4 omega l square c square 4 4 omega omega l c l square c square get cancelled omega omega c square c square gets cancelled 2 to the 4. So, term remaining is 1 upon l c minus r square upon 2 l square is it. Now, if we take square root it will be becomes omega is equal to under root 1 upon l c minus r square upon 2 l square is it. And yes omega is equal to 2 pi f. So, therefore, f is equal to under root 1 upon l c minus r square upon 2 l square into 1 by 2 pi is it. So, this is f c frequency at which voltage across capacitor is maximum and voltage across capacitor is maximized because of this differentiation when we differentiate v c square with respect to omega and we keep the value or when we equate the numerator to 0 the v c will be maximized. v c if v c is maximized v c square is also maximized to reduce the mathematical calculation we have taken the square of v c and whatever the term is coming we have equated or we have taken the differentiation from that and we have equated it to 0. And for that we have used u by v root is it. So, f c is equal to 1 upon 2 pi under root 1 by l c minus r square upon 2 l square this is the value of frequency at which voltage across capacitor is maximum is it. So, this is the derivation for f c in which the frequency is this and it is in hertz. So, this is all related to the derivation references while preparing this lecture I have used circuit theory analysis and synthesis by H. Just like this we can write down the formula or we can derive for frequency at which voltage across inductor is maximum that we will see in next lecture. Thank you.