Current Calculation Problem (Probably Simple)

[Iscariot]

A really basic circuit problem which I just don't understand.

In the attached figure, C is a cell of e.m.f. 2V and internal resistance 1 Ohm. Calculate:

a) the current through A
b) the current through B
c) the p.d. between the terminals of the cell

I really don't understand how to calculate the current through A. If someone could just help me with that then I'd appreciate it, I'm hoping that from someone pointing out what's probably obvious in part A, I'll be able to do B and C by myself.

 

[Iscariot]

Never mind, I solved my own question. Sorry about this, it took me hours.

 

[quicknote]

First of all, it is important to understand the basic principles of electricity in order to solve this problem. In this circuit, the cell (C) acts as a source of electrical energy, providing an electromotive force (e.m.f.) of 2V. However, the cell also has an internal resistance of 1 Ohm, which means that some of the energy is lost as heat within the cell itself.

To calculate the current through A, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the resistance is the sum of the internal resistance of the cell (1 Ohm) and the resistance of the wire connecting A and B. Let's call this resistance R1.

Therefore, the current through A can be calculated as I = V/(R1 + 1 Ohm). We know that the voltage provided by the cell is 2V, so we just need to find the value of R1 in order to solve for the current.

To find the value of R1, we can use Kirchhoff's Voltage Law, which states that the sum of the voltage drops in a closed loop circuit must equal the voltage supplied by the source. In this case, the loop is A-B-C-A, and the voltage drops are V1 (between A and B) and V2 (between B and C).

Since we know that the voltage drop across the cell is 2V, and the voltage drop across the internal resistance is 1V (due to Ohm's Law), we can set up the following equation:

2V = V1 + V2 + 1V

We also know that the voltage drop across a resistor is equal to the current flowing through it multiplied by its resistance (V = IR). Therefore, we can rewrite the equation as:

2V = I*R1 + I*1 Ohm + 1V

Substituting the value of I from our previous equation, we get:

2V = (2V/(R1 + 1 Ohm))*R1 + (2V/(R1 + 1 Ohm))*1 Ohm + 1V

Simplifying and rearranging, we get:

2V = 2V + 2V*R1/(R1 + 1 Ohm)

Solving for R1, we get:

R