5

2. For the circuit of Figure P3.2, the output is across the 2Q resistor: Find a state-space representation. [Sec- tion: 3.4] ae 103 sUuuu 1 H2 F2 Qip(t)FIGURE P3.2...

Question

2. For the circuit of Figure P3.2, the output is across the 2Q resistor: Find a state-space representation. [Sec- tion: 3.4] ae 103 sUuuu 1 H2 F2 Qip(t)FIGURE P3.2

2. For the circuit of Figure P3.2, the output is across the 2Q resistor: Find a state-space representation. [Sec- tion: 3.4] ae 103 s Uuuu 1 H 2 F 2 Q ip(t) FIGURE P3.2



Answers

For the circuit shown in Figure $\mathrm{P} 28.32$ , we wish to find the currents $I_{1}, I_{2},$ and $I_{3}$ . Use Kirchhoff's rules to obtain equations for $(a)$ the upper loop, (b) the lower loop, and (c) the junction on the left side. In each case, suppress units for clarity and simplify, combining the terms. (d) Solve the junction equation for $I_{3}$ . (e) Using the equation found in part (d), eliminate $I_{3}$ from the equation found in part (b). (f). (f). (f) Solve the equations found in parts (a) and (e) simultaneously for the two unknowns $I_{1}$ and $I_{2}$ . (g) Substitute the answers found in part (f) into the junction equation found in part (d), solving for $I_{3} .$ (h) What is the significance of the negative answer for $I_{2} ?$

I have friends for ideally operational amplifier. Re positive is called to be negative. I posited is going to high negative. Start to zero. Let us writing nodal equation for inverting terminals, we can write way to minus B minus upon our one must be not minus B, negative. All hard minus are negative. Be positive. It's called to fever. Also be negative. So every question would be B two minus V. Evil upon our one. Tell us be not minus people upon our to start zero, so be not. You will get one plus r upon our one into the river minus R. Two upon are born into video. That's all. Thanks for watching.

Hi in the given problem there are the two resistors having resistance values as Arvin is equal to 11.0 ohm and are too which is equal to 7.0 home. These two resistors are joined in parallel across a battery which is having an IMF value of 50.0 world. No, in the first part of the problem, we have to find the equivalent resistance of this parallel combination here are one is 11.0 form and our two is seven point double zero home. So in parallel, the formula used to find equivalent resistance is the inverse of equivalent resistance is given by the addition of inverse of the individual raises stores. So plugging in alone values here, this is one by 11 plus one bite seven. Who's Elsie? Um is 77. So this is seven plus 11 means this is 18 by 77. Therefore, equivalent resistance of this combination will be 77 by 18 means it will come out to be poor point 28 home which becomes the answer for that first part of this problem. Now, in the second part of the problem, we have to find the total current passing through the circuit means the current coming out of this battery. So using own slaw, That current will be given by the ratio of the IMF light with the equally violent resistance. So here it will be given by I is equal to 50.0 volt divided by equivalent resistance, which was 4.28 oh. So finally this current passing through the circuit will come out to be 11.7 NPR which becomes the answer for the second part of this problem. Then in the third and the fourth part of the problem, we have to find the currents passing through the resistors individually through our one suppose it is Ivan and through our to suppose this is by two. So using home slow again and the concept as potential remains the same in parallel. So bold the resistors, we will be having same 50.0 volt. It grows them. Therefore, I even current passing through our own using home slope will be given us E by our one. Yes, this is 50 world divided by 11 0, which finally we come out to be Ivan is equal to 4.5. Bye and Fear. And similarly in the fourth part of the problem, the current passing through resistance are too. That will be given by E by our two means 50 volts divided by 7.0 Oh, so finally this current year will come out to be 7.15 M p r. Answer for the fourth and the last part of this problem. Thank you.

Hi the given problem. The three resistors are in parallel here. This is our one then this is R. Two and the third one. R three all are in parallel joined with a battery providing an IMF having a value of 10.0 world. The value of this resistance R one is 2.0 home. There are two is five 100 home. And finally this artery is 8.0 on the total current coming out coming out of the battery. If it is assumed to be I, then the current passing through our one is Ivan. That passing through art will be I too. And finally through. No. In the first part of the problem we have to find the current passing through the resistance are too for which the concept which will be used here is the potential remains the same in parallel. So the potential will be same across all these three days and stores. So in the first part of the problem, current passing through resistance art will be I took and that will be given by own slaw as I physical to E by our two since this is 10 world by five homes. So finally this I too comes out to be 2.0 ampere answer for the first part. Then in the second part using the same concept and seen home slope the current passing through resistance R three which is I three will be given by E by our three means 10 walls again but divided by 8100 own. So this time this current I treat here comes out to be one point 25 NPR which becomes the answer for them. Second part of this problem and then the third part of the problem, we have to find current passing through the first resistor R one. So that will be Ivan is equal to 10 walt. Again divided by R one which is 2.0 Oh so finally this Ivan here comes out to be 5.0 and pure. Which is answer for them. Third part of this problem. Now in the fourth part of the problem we have to find the total current coming out of the battery which will be given by I is equal to I won unless I do plus I three. All these three currents passing through the various resistors will be joined to give so it becomes 5.0 plus 2.0 plus 1.25 M. Pierre. So it becomes I is equal to 8.25 ampere which is the answer for them. Fourth part of this problem. Finally, in the last part of the problem, we have to find equivalent resistance of this combination. So using arm slot, equivalent resistance will be given by net mff light, divided by net current coming out of the battery. So on net IMF applied, this is stan walt again, and the net current is 8.25 anthea, which we have found in the fourth part of this problem. So finally, equivalent resistance here comes out to be 1.21 home, which is the answer for that fifth and the last part of this problem. Thank you.

Border. Given practical subject, we can write the equation from modal analysis, deacon, not for will be different. So equation for notes, one to empty, we will is jailbird Leaving up one ft. Let's be too into Man by three. Has been vital has been by four minus the three. Into one by trees, conclude Europe for two weekends. Right marinos. We even into one upon one, Me too into open mercury B three into 1 month to respond by two. And that's when my son is part two solving an upset of situations. We can people get even strangled due to export what private boat and we G is 7.3 Equal mm. Thanks for watching it.


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