Can someone explain my answer to me? I want to understand it.

[frozenguy]

Homework Statement

Hi physics forums, thanks for responding if you do
Use Ohm's law and Kirchhoff's current law to determine the current I₁ in the circuit of Figure--

The Attempt at a Solution

So I'm trying to fully understand what is going on here..

Besides not getting where the voltage comes from without a voltage source; where is the 100volts? Its the drop over the equivalent resister R_p, I know that. So I guess there is 0 volt on the ground side (-?) of the current source and 100 volts on the opposite (+?) side.

I used 100 volt to find the current through R1, but its 100 volts through R2 also? How is it 100 volts through both?

 

[Zryn]

You have a current source in parallel with a resistance (R1 or R2 or RP) which can be modeled as a voltage source in series with a resistance (maybe not learned that yet? you don't need to do it to calculate the voltages anyway).

Either way, you can have a voltage with no current flowing (terminals of a battery) but you can never a current flowing with no voltage, thus to have a current source, there must be a voltage somewhere driving the current.

Ohms Law shows us this, in that for any current flowing through a resistor there will be a voltage V, as you have calculated.

Yes the (-) is 0 and the (+) is 100, as per Kirchoffs Voltage law around any of the loops.

Are you aware of the general rule that voltages in parallel are equal and current in series is the same everywhere?

 

[frozenguy]

You have a current source in parallel with a resistance (R1 or R2 or RP) which can be modeled as a voltage source in series with a resistance (maybe not learned that yet? you don't need to do it to calculate the voltages anyway).

Thevenin equivalent? We were lectured on it but I haven't studied it. And its after this section in the book.

Either way, you can have a voltage with no current flowing (terminals of a battery) but you can never a current flowing with no voltage, thus to have a current source, there must be a voltage somewhere driving the current.

Ohms Law shows us this, in that for any current flowing through a resistor there will be a voltage V, as you have calculated.

Yes the (-) is 0 and the (+) is 100, as per Kirchoffs Voltage law around any of the loops.

Are you aware of the general rule that voltages in parallel are equal and current in series is the same everywhere?

Isn't V=I1R1 the voltage drop over the resister? Meaning it has less voltage after it goes through? So how can the voltage drop be 100 for each if there is only 100 to begin with?

 

[frozenguy]

Did I do this right? Finding the power delivered by the current source?

My teacher says its negative, but I see positive voltage on the positive side of the current source, and its a positive current. So how does he get negative?

 

[DeeNos]

Sure, I can explain your answer to you. Let's break it down step by step.

First, let's address your question about the voltage source. It is not explicitly shown in the circuit, but it is implied that there is a voltage source connected to the circuit. This is necessary for the circuit to have a current flowing through it.

Next, you correctly identified that the 100 volts is the voltage drop across the equivalent resistor Rp. This is because according to Ohm's law, the voltage drop across a resistor is equal to the current flowing through it multiplied by the resistance. In this case, the current flowing through Rp is the same as the current flowing through R1 and R2, so the voltage drop is the same.

Now, let's look at Kirchhoff's current law. This law states that the sum of the currents entering a node (or point) in a circuit must equal the sum of the currents leaving that node. In this circuit, the current I1 is entering the node where R1, R2, and Rp are connected. This means that the sum of the currents leaving that node must also be equal to I1. Since R1 and R2 are in series, they have the same current flowing through them. Therefore, the current through R2 is also equal to I1.

I hope this helps to clarify your understanding of the circuit. If you have any further questions, please don't hesitate to ask. Remember, understanding is key in science, so it's great that you are seeking to fully understand this problem. Keep up the good work!