What value of R in the circuit of Fig. 2.109 would cause the current source to deliver 800 mW to

What value of R in the circuit of Fig. 2.109 would cause the...

Key Concepts

Circuit Analysis Techniques

Circuit analysis techniques involve applying principles such as Ohm's Law, Kirchhoff's Voltage Law, and Kirchhoff's Current Law in order to simplify and solve circuits. These methods are used to calculate unknown parameters—such as resistor values—that achieve desired outcomes like specific power delivery.

Ohm's Law

Ohm's Law, expressed as V = IR, is fundamental in circuit analysis as it relates voltage, current, and resistance. This relationship is crucial for understanding how changes in resistance affect voltage drops and, consequently, the power delivered by a current source in a circuit.

Current Sources

A current source is a circuit element that supplies a constant current to a circuit regardless of the voltage across it (within its operational limits). Understanding current sources is key to analyzing how current is distributed and how voltage develops across the load elements connected to them.

Resistive Power Dissipation

Power dissipation in resistive components occurs when electrical energy is converted into heat. This is quantified by equations such as P = I²R, where P is power, I is current, and R is resistance. These calculations are essential for determining how much power is delivered to or consumed by resistors in a circuit.

Transcript

00:01 In this problem of dc circuits we have given a figure let me show you first.

00:07 So this is the circuit and we have to find the value of r in this circuit such that it would cause the current source to deliver 800 mil watt to the resistor.

00:22 So here it would be delivering 800 milli watt.

00:28 So here v is equal to say v is equal to 30.

00:34 Millie ampere this would be actually current so here this would be current is equals to 30 millie ampere so here's power is given why i multiplied with v since we don't know value of v so v is replaced with i r so this would be i square multiply with r equivalent so here we have value of i and we have to find the value of this r so first we would find the value of r equivalent so here this would be equal with equated with 800 mill watt so i is equal to 30 so this would be equals to 30mphal whole square so this would be 30 multiplied with 10 to the power minus 3 whole square multiplied with our equivalent is equal to 800 multiplied with 10 to the power minus 3 watt so from here this is would be r equivalent is equal to this would be 800 divided with this would be 30 square is equals to 900 and this is multiplied with 10 to the power minus 30 this is minus 3 and this would be in oom so from here 0 to 0 is cancelled out and we have r equivalent is equals to 8 divided with 9 and multiplied with 10 to the power 3 oom so here first we have to convert this pi into y so this would be the new circuit like this i'm reading so i'm writing it with the new color so this would be your new and here so we have to find the value of r1 r2 and here this would be again a resistor so this would be r3 although all would be the same so here we are converting from delta to y so from delta to y, r1 here in this case is equal to r2 and this is equals to r3...

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