[Skie]

Guys help me out with these 7 questions

[SmG Abi]

I'm way too stupid to know any of that

[Skie]

Guys help me out with these 7 questions

[SmG Conan]

The first one I = V/R. There is 2 voltages sources on series but they have different polarities, reduce them together and you get one voltage source of 14v. Going the way of the current. you reduce r3 and r1 since they are in series. Then just plug in: I = 14v/R3+R1

The second one I am not sure entirely but if you follow the circuit diagram v1 should be -10v since +10v-20v and v2 -6v ?????? I am not sure again.

Next one is R = V/I. 170v/5A since it outputs 5a to loads and it has a 170 no load output. You should get Rint with that.

Next one is I = V/R. 5 resistors connected in series and a 120v source. I = 120v/(30*5). Power is P = V*I, with the precious answer you can just plug in P = 120v * I. They are all connected in series and with the same resistance value so the power is equally distributed. Voltage drop V = I * R. From previous answer V = I * 30ohms. All the drops are the same. If the bulb dies and the filament opens the whole circuit breaks since they are connected in circuit.

2 Resistors in series total up to one resistor on 4kohms . 2 sources in series total to 8v. Is = Vs/Rt. Is = 8v/4kohms. V1 = 8v/1kohms and Va is Vs - V1 since in the resistor is a voltage drop.

Vs = E1 + E2. Rt = R1 + R2. Vs = Is/Rt. Now plug in.

Last one is Vab= Vb - Va. Vb = E1. Find Rt then Vs then E2 Then do voltage drop on the 200ohms resistor and you will find Va. I am not sure about this one either.

[Skie]

The first one I = V/R. There is 2 voltages sources on series but they have different polarities, reduce them together and you get one voltage source of 14v. Going the way of the current. you reduce r3 and r1 since they are in series. Then just plug in: I = 14v/R3+R1

The second one I am not sure entirely but if you follow the circuit diagram v1 should be -10v since +10v-20v and v2 -6v ?????? I am not sure again.

Next one is R = V/I. 170v/5A since it outputs 5a to loads and it has a 170 no load output. You should get Rint with that.

Next one is I = V/R. 5 resistors connected in series and a 120v source. I = 120v/(30*5). Power is P = V*I, with the precious answer you can just plug in P = 120v  * I. They are all connected in series and with the same resistance value so the power is equally distributed. Voltage drop V = I * R. From previous answer V = I * 30ohms. All the drops are the same. If the bulb dies and the filament opens the whole circuit breaks since they are connected in circuit.

2 Resistors in series total up to one resistor on 4kohms . 2 sources in series total to 8v. Is = Vs/Rt. Is = 8v/4kohms.  V1 = 8v/1kohms and Va is Vs - V1 since in the resistor is a voltage drop.

Vs = E1 + E2. Rt = R1 + R2. Vs = Is/Rt. Now plug in.

Last one is Vab= Vb - Va. Vb = E1. Find Rt then Vs then E2 Then do voltage drop on the 200ohms resistor and you will find Va. I am not sure about this one either.

Thanks man 

[SmG Patriciocia]

Wow these questions require too much brain power.

[ItsOrangeRBX]

Dang, that looks very hard.

[SmG Shintaro]

I need subtitles

[«SmG» Cloud]

I know an EE do you still need this done?