I'm hoping you can help me calculate I1 and I2. I already have the following answers which I have already calculated:
I have replace R2, R3 and R4 with a single resistor called Rt by:
Rs = R2 + R3
Rs = 8 + 4
Rs = 12 Ohms
1/Rt = 1/Rs + 1/R4
1/Rt = 1/12 + 1/12
Therefore,
Rt = 6 Ohms
To calculate It, I added Rt and R1 together to give me a total circuit resistance of 12 Ohms, and from the picture, you can see the voltage is 12v.
So using Ohms law, I = V/R = 12/12 = 1 A
I am unsure of which route to go down next. I have calculated the voltage drop across R1 as 1A x 6 Ohms = 6V, but I am unsure how to calculate (show working) I1 and I2.
I think they are going to be the same currents, as the resistances are both 12 Ohms, so I think I1 and I2 are both 0.5A in this instance, as reading Kirchhoff's Law, it says the sum of the currents are the total of the total current, so in this case 0.5A + 0.5A = 1A which I calculated earlier.
But what if I2 was 24 Ohms for example, which formula would I need to use? I'm struggling the show my workings for I1 and I2.
Best Answer
Now that you have the value of $I_T = 6$ A, you can apply Kirchhoff's Law to calculate the voltage drop in the resistance you call $R_S$:
$$ -12V + 6\Omega I_T + I_1 R_S = 0 \quad\Rightarrow\quad I_1 = (12V - 6\Omega\times 1A) / 12\Omega = 1/2\;A $$
You could do the same for $I_2$, but an easier way is to realize that $I_T = I_1 + I_2$ therefore $I_2 = 1/2\;A$