Electric eqs q

Working and diagram of question
I colour coded to make it clear to myself that i3=i1+i2

(edited 1 month ago)

Reply 2

Original post by mosaurlodon

Working and diagram of question
I colour coded to make it clear to myself that i3=i1+i2

I am unsure what your convention is for writing positive and negative potential differences.
Based on your right mesh/loop (yellow coloured and green coloured lines), you move clockwise through the cell, which is a potential drop, it should be -V2 and based on your assigned current direction, I have
-V2 - I2(1000) - I3(15) = 0

Reply 3

Ohhhh thank you!
So in my diagram If I flipped the right cell vertically ie

Would my original eqs be correct or would it be -V2 = I2(1000) + I3(15)

Reply 4

Original post by mosaurlodon

Ohhhh thank you!
So in my diagram If I flipped the right cell vertically ie

Would my original eqs be correct or would it be -V2 = I2(1000) + I3(15)

What is the actual drawing of the circuit?

Reply 5

The original q comes from part b; https://isaacphysics.org/questions/two_cs_three_rs?board=07259f72-2a30-45f9-8a89-35627f34d9a5&stage=a_level
I just worked out the voltages of each capacitor and made a simplified diagram.

Reply 6

Original post by mosaurlodon

Ohhhh thank you!
So in my diagram If I flipped the right cell vertically ie

Would my original eqs be correct or would it be -V2 = I2(1000) + I3(15)

If you want to flip the polarity of V2, you are changing the question to suit your written equation and you would still get wrong I3.

Reply 7

Then I have a further q - for part b which polarity for both the capacitors do you take, since you have to assume they act like cells for kirchoffs laws?

Reply 8