So, I did this practical where the circuit has an x and y terminal that connects to a box that has a three resistors and a diode.
There are places you can connect the circuit, A, B and C.Illustrated here, https://i.imgur.com/0UBelxh.jpg, the top is the circuit and the box is the box. So I did all possible connection and collated my results, here, https://i.imgur.com/6XoSZ4j.png.
I'm trying to ultimately find the value of R3, the voltage across the conducting diode was given to me and is 0.70 V. I'm to consider to voltage between A and C when A is positive and calculate:
1. The voltage across R3.
My table says 4.48 V, so I just subtracted 0.7 V from it.
2. The current through R2 and R1.
I think when X is B and Y is A will give me the current for R1 at 22.19 mA and when X is C and Y is B gives me the current for R2 at 88.9 mA.
3. The value of R3
I think the law is current splits up in each strand, so I can use the combined current of R1 and R2 and divide the voltage in question 1 by it and get the value of R3.
I'm not 100% certain on question two and three, I'd appreciate if anyone could help.
There are places you can connect the circuit, A, B and C.Illustrated here, https://i.imgur.com/0UBelxh.jpg, the top is the circuit and the box is the box. So I did all possible connection and collated my results, here, https://i.imgur.com/6XoSZ4j.png.
I'm trying to ultimately find the value of R3, the voltage across the conducting diode was given to me and is 0.70 V. I'm to consider to voltage between A and C when A is positive and calculate:
1. The voltage across R3.
My table says 4.48 V, so I just subtracted 0.7 V from it.
2. The current through R2 and R1.
I think when X is B and Y is A will give me the current for R1 at 22.19 mA and when X is C and Y is B gives me the current for R2 at 88.9 mA.
3. The value of R3
I think the law is current splits up in each strand, so I can use the combined current of R1 and R2 and divide the voltage in question 1 by it and get the value of R3.
I'm not 100% certain on question two and three, I'd appreciate if anyone could help.