Figure 9.81 shows a parallel combination of an inductance and a resistance. If it is desired to

Figure 9.81 shows a parallel combination of an inductance...

Key Concepts

Frequency Dependent Reactance

The reactance of inductors and capacitors depends on the frequency of the AC signal, with capacitive reactance decreasing and inductive reactance increasing as frequency rises. Recognizing this behavior is key in designing circuits that adjust the overall impedance at particular frequencies by canceling out the reactive components.

Series and Parallel Impedance Combinations

Understanding how impedances combine in series and parallel configurations is crucial for circuit analysis. In series connections, impedances add directly, while in parallel configurations they combine according to reciprocal relations. This knowledge is applied when modifying circuits with additional reactive elements to achieve specific impedance characteristics.

Resonance in RLC Circuits

Resonance occurs when the reactive effects of inductors and capacitors cancel out at a specific frequency, resulting in a purely resistive impedance. This principle is essential for tuning circuits to operate at desired frequencies by setting the conditions for reactive cancellation.

Complex Impedance

Complex impedance represents the combination of resistance (real part) and reactance (imaginary part) in an AC circuit, allowing engineers to account for both magnitude and phase differences between voltage and current. This concept enables analysis and design of circuits where the behavior of resistors, inductors, and capacitors is interdependent on frequency.

Transcript

00:01 A student in this problem a parallel combination of inductance and resistance is given.

00:06 So here an inductor l and a resistor r is connected.

00:15 And it is desired to have connect a capacitor c in series with this circuit so that our net impedance of the circuit will be resistive and it is at the frequency of 10 megahertz.

00:32 Okay frequency of 10 megahertz okay and the value of inductor is 20 micro henry and the register is 300 om and the register the value of capacitor is c so we have to find out the value of capacitor for the circuit will circuit will be purely resistive at the frequency of 10 megahertz okay so now first we have to calculate the total impedance of the circuit z -in.

01:03 So z input will be here the resistor and the inductor are connected in parallel, both are connected in parallel in series with the capacitor c so the net impedance the net input impedance is is 1 upon 1 upon j omega c plus are parallel j omega okay so so z input equal to z input equals to 1 upon j omega c plus plus by calculating the parallel combination it give j omega r l upon r plus j omega j omega l okay so now z input equals to by separate by taking iota in numerator term minus iota upon omega c plus by multiplying the complex conjugate in this term it give j omega r l multiply r minus iota omega l upon it give r r square plus omega square l square so now z input equals to j input equals to minus j upon omega c plus by multiply this function j omega r square j omega r squared l and here minus sign and iota into iota is square is also minus 1 so we put plus plus omega square l square r upon r square plus omega square l square equation number first okay so now here we apply the impedance concept okay so in impedance concept in this circuit the net impedance will be resistive whenever we connect our capacitor so in that in that condition the imaginary component of the imaginary component of z input equals to 0 when circuit will be will be resistive okay circuit will be resistive so we put the imaginary component in this function will be 0 okay so in this here this is imaginary function and here this is our imaginary function okay omega r square l upon i square plus omega square l okay so here minus 1 upon omega c plus omega r square l upon r square plus omega square l squared put equal to 0 okay now by calculating this value 1 upon omega c equals to omega r square l upon r squared plus omega square l if you solve this function for the value of capacitor then our capacitor c will be r square plus omega square l square upon omega square l r square okay so this is the value of this is the value of capacitor okay this is the value of capacitor and we know that the frequency will be or frequency will be 10 megahertz okay or frequency will be 10 megahertz f equals to 10 megahertz okay so f equals to 10 k power 6 it gives 10 k power 7 now we have to calculate the value of capacitor we we have to calculate the value of capacitor...

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