 Some friends today will see frequency for maximum voltage across inductor. So, generally there are three different stages of variation of frequency at which voltage is maximum across inductor across resistor across capacitor yes. So, when frequency is at resonant frequency voltage is maximum across resistance or resistor because magnitude of voltage across inductor and capacitor is same at resonance. So, both cancels each other both are same in magnitude and opposite in phase. So, both cancels each other and the whole voltage will appear across resistor. So, voltage is maximum across resistor in case of resonance. Now, the question is what is the frequency what is the value of frequency at which voltage is maximum across inductor that we are going to find out ok. Learning outcome students will be able to demonstrate how to find frequency at which voltage across inductor is maximum in case of series resonance. And before starting with the derivation you should recall equations for voltages, equation for current, equations for impedance, magnitude of impedance, variation of frequency with respect to impedances or what is the effect of variation of frequency on impedances all those things. Now, we will start with the derivation yes. So, we will start with the frequency for maximum voltage across inductor. So, we have to write down the equation for voltage across inductor or equation for voltage V L. It is represented by I into X L yes, X L can be also represented by omega into L reactance is equal to omega L and in case of resonance current is represented by V by Z because there is resistor as well as reactive part. So, it is represented by impedance. So, here Z represents the magnitude of impedance therefore, V L is equal to V into omega into L upon R square plus omega L minus 1 upon omega C bracket square is it. Now, V square on both sides. So, we will get V L square is equal to V square omega square L square upon R square plus omega L minus 1 upon omega C bracket square. We will further simplify this as V L square is equal to V square omega square L square upon R square plus. Now, we cross multiply this, it will be omega square L square C square minus 1 bracket square upon omega square C square. Now, V L square is equal to V square omega square L square. Now, again we cross multiply this, it becomes omega square R square C square plus omega square L C minus 1 bracket square. It is not L square C square sorry it is L C this term divided by omega square C square. Now, this term we will go to the numerator. So, V L square is equal to V square omega raise to 4 L square C square upon omega square R square C square plus omega square L C minus 1 bracket square. Yes. Now, what we have to do is we have to differentiate V L square with respect to omega to maximize the voltage and equate the numerator term to the 0. So, again it is in the form u by v. So, again for differentiation we have to use u by v rule V du minus udv upon v square is equal to 0. Differentiating V L square with respect to omega and equating numerator term to 0 d V L square by d omega equal to 0. I will directly write down the term after simplification the terms are like this 2 into omega square L C minus 1 sorry 2 in bracket omega square L C minus 1 minus omega square R square C square equal to 0. So, this becomes 2 omega square L C minus 2 minus omega square R square C square equal to 0. If we take this 2 to the RHS it becomes 2 omega square L C minus omega square R square C square is equal to 2. Now, from this I will take the term omega square common. So, it become 2 L C minus R square C square is equal to 2. Now, I take this term to the RHS it becomes omega square is equal to 2 upon 2 L C minus R square C square. Now, what we do is we divide numerator and denominator by 2. Now, it becomes omega square is equal to 1 upon L C minus R square C square divided by 2. After dividing numerator and denominator by 2 we will get term like this square root on both sides. Now, take square root on both sides it becomes omega is equal to 1 upon under root L C minus R square C square upon 2. Yes, and if you put omega is equal to 2 pi f here then f L becomes 1 upon 2 pi under root L C minus R square C square upon 2. f L is equal to 1 upon 2 pi under root L C minus R square C square upon 2. Now, if you take L C term outside or from that bracket f L is equal to 1 upon 2 pi under root L C under root 1 minus R square C upon 2 L. And 1 upon 2 pi under root L C is resonant frequency and upon 2 pi under root L C is equal to f 0 resonant frequency. Therefore, f L is equal to f 0 upon 1 minus R square C upon 2 L. So, this is the formula for frequency at which voltage across inductor is maximum. So, either you can use this formula or you can use this just what I have done is I have taken term L C common and instead of 1 upon 2 pi under root L C I have used term f 0 that is resonant frequency. So, this is all related to the frequency at which voltage is maximum across inductor. So, while preparing this video lecture I have used circuit theory and analysis by H. Chakraborty Zanpatraya publication. Thank you.