For the circuit below, calculate (a) the voltage V 1 shown, and (b) the current through the 6-Ohm resistor.

Hey guys. Hope you try this practice problem on your own. If you haven't, you should pause the video and give it a shot now. I'm gonna keep rolling. So we're looking for voltage v 1, which is this guy over here. This voltage is given. And the currents through the 6 ohm resistor. 6 ohms right here. I want this current. So let's call this I one and say that I want v one as well as I one. So that's what I'm looking for. These 2 guys. Cool. How are we gonna solve these problems? Well, the general steps are to 1, label a lot of stuff and 2, to write a lot of equations. As many equations as you need to solve it. So we're gonna label junctions, loops, currents, and finally currents. And finally we are going to put pluses and minuses on resistors and batteries. Alright? And we're gonna write equations to solve it. So let's get started. So junctions are easy. There's a junction here a and there's a junction here b. Loops, we can pick any spot. I'm gonna pick this spot right here and we're gonna loop this way which is going to allow us to go with, the eye in the direction of these two currents right here. By the way, I don't actually know that this current is in this direction, but I am assuming that it is and that's fine. Right? And then I 2 I'm sorry, loop 2, we're just gonna go this way as well. Loop 2. Okay? Remember the directions of the loops are arbitrary. The loop directions are just the sequence in which you're going to jump around the circuit the circuit elements as you're writing the equations. So I've labeled our the junctions and the loops. What about currents? So these currents are labeled in a direction. We're gonna call this I 2 and then there's an i3 here. The positive of the battery is this way so I'm just gonna assume that the i3 is going in this direction. K. So I got I three right here and then I got I one right here. So I got my currents labeled and I can label pluses and minuses. I can put little pluses everywhere. So there's a plus on the positive side of the battery, so a negative on the other. Plus on the positive, which is the long side of the battery, negative on the other. The pluses and the resistors go where the current enters. So the current enters here. The current enters here. So that gets a plus. And then this I one enters right here. So that gets a plus. So I'm done labeling all the pluses and minuses. Now we're gonna write equations. We're first gonna write the junction rule or the junction equation because it's just simpler. So at junction a, I can write that currents in equals currents out. Currents in are I one plus I three and that equals currents out I two. But I already know I 2, so I can just I can just say that this equals 4. K. So I 1+i3 equals 4. Alright. If I write junction b, I'm gonna get the same thing. I'm gonna do it real quick but you've seen this a few times probably where the equation ends up being the same. So I have this 4 amp going in. What else? If you backtrack the I one over here, I one looks like this. And if you backtrack the I three, I three looks like this. So you can see that 4 goes in and then I one plus I three goes out. This equation is the same as that one. It's not an original equation so it's not really useful for us. Anyway, we got one equation so far. And let's write loop equations now. So loop 1 is gonna we're gonna say that the sum of all voltages equals a lot of crap which equals 0. And let's start jumping. So we're gonna start over here and we're gonna go around this thing. Right? So So we're gonna travel around this thing. I'm going to jump from a negative to a positive. Therefore, this is going to be a positive 12 volts. And then here I'm gonna jump from a negative from a positive to a negative. So this is gonna be negative, the voltage of the 6. Voltage of a resistor is given by Ohm's law, v equals ir. So here I'm gonna put ir. The current is I one and the r is 6. We're gonna keep walking around here and then we're gonna jump from a positive to a negative. So So this is gonna be negative. This is also resistor, so it's IR again. And the current here, we know it is a 4 and the the current here is a 4 and the resistance, the resistance is also a 4. K? So if you write this out, you're gonna get 12 minuteus 6i 1. 12 minuteus 6i I one minus 16 equals 0. I have 12 minus 16, which is negative 4. So negative 6i1-4 equals 0 or I one equals the 4 is gonna go to the other side as a positive and then divided by negative 6. So this is gonna be negative 0 0.67 amps. Okay? Negative 0.67 amps. I got a negative. What does that mean? Remember, the negative just means that our assumed direction for I one was actually incorrect. It doesn't change the results. You don't have to start over. You just know that I one, the actual direction of I one is that way. You're gonna keep solving the problem. In fact, you're gonna keep using the negative 0.67 because now you gotta stick with it. Now you gotta be consistent. Cool? Alright. So l 2, loop 2. I'm gonna move this a little bit so I don't block. Loop 2, the sum of all voltages on loop 2, lots of stuff, equals 0. K? Loop 2 is gonna start over here in point b and we're gonna go, in this direction now. We're gonna jump from a from a negative to a positive. So this is gonna be a positive voltage. We're crossing a resistor. So it's gonna be I times r. I know INR 4 and 4. Gonna keep going around here. We're now gonna cross this other resistor but now I'm going from, again going from negative to positive. So this is a positive. Again, IR because you're crossing resistor. And lastly, we're gonna jump this voltage one here from positive to negative. So this is gonna be negative v one. I forgot to put these numbers here. This is the current is I 3, which we actually don't have. And the the resistance is a 3. Right here. I 3 and a 3. Cool. So if you look at this equation, let's actually clean this up a little bit. So 16+3i3 minusvone equals 0, or I can just move the v one to the other side. V one is what we're looking for. But I can't solve v one unless I have I 3. Do I have I 3? Not yet. But I can probably calculate it. Whenever you want a current, you should look at your junction equation because it's gonna have the all the relationships between all the currents. And notice that right here we calculated I one. So we have I one, which means we can find I 3. So let's go off to the side over here and I'm just gonna take a little little detour here. And we're gonna say I one plus I three equals 4. I know I one so I can find I three. I three is 4 minus I one. I 1 is 0.67. So 4 minus, be careful here, 0.67. Minus minus becomes a plus plus. Right? Two negatives make a positive. So this gives you 4.67 amps. That's your I three. K? I 3 is 4.67 amps. Whoops. Can't see that. Sweet. So now we can plug that guy in here and we end up with v 1 equals 16, plus 3 times 4.67 equals so if you do all of this, you end up with 30 volts. 30 volts. Okay? And we're done. If you go back to the beginning of the problem, you see that the things we were asked for were the voltage v one, which we just found. That's 30 volts. Right? And we were looking for the currents through this resistor I 1. And I 1 is 0.67. 0.67. K? Now if you're answering the question, if if your professor asks you, what is the current y one? You should use the positive version of that current. It is 0.67. The negative over here just means that when you were solving the problem, you got the direction wrong. That's cool. That's not bad. And if he asks you the direction, you or she asks you the direction, you would say that it was clockwise on this circuit right here. Clockwise on that loop. Cool? That's it for this one. Hope it makes sense. Hope you got it. Let's keep going.

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27. Resistors & DC Circuits

Kirchhoff's Loop Rule

Physics

27. Resistors & DC Circuits

Kirchhoff's Loop Rule

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Multiple Choice

For the circuit below, calculate
(a) the voltage V₁ shown, and
(b) the current through the 6-Ohm resistor.

i_6Ω = 4.67 A, V₁ = 1.8 V

i_6Ω = 4.67 A, V₁ = 24 V

i_6Ω = 0.67 A, V₁ = 26 V

i_6Ω = 0.67 A, V₁ = 30 V

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Verified step by step guidance

Identify the components in the circuit: a 6Ω resistor, a 4Ω resistor, a 3Ω resistor, a 12V battery, and a voltage source V1.

Apply Kirchhoff's Voltage Law (KVL) to the loop containing the 6Ω resistor, 4Ω resistor, and the 12V battery. The sum of the voltage drops across the resistors should equal the battery voltage.

Calculate the voltage drop across the 4Ω resistor using Ohm's Law: V = I * R, where I = 4A and R = 4Ω.

Determine the voltage drop across the 6Ω resistor using Ohm's Law: V = I * R, where I is the current through the 6Ω resistor, which needs to be calculated.

Apply KVL to the loop containing the 3Ω resistor and V1. Use the known current through the 3Ω resistor to find the voltage drop across it, and then solve for V1 using the equation V1 = total voltage - voltage drop across the 3Ω resistor.