Help with Kirchoff's law problem?

https://postimg.cc/DJ8HW4Sy

I don't have the markscheme so I am unsure if this is correct:

9v = I1×350 → I1 = 9/350

6v = I2×350 → I2 = 6/450 = 1/75

So via kirchoff's 2nd law I1+I2 = I3 = current down the 50 ohm resistor, hence power dissipated across it = (9/350 + 1/75)²x50 ≈ 0.0762w (3sf)

Reply 1

lordaxil

This is not correct - you cannot consider the current flowing through two loops independently, as the SAME current must pass through the 50 Ohm resistor in both cases. Try writing down the equations for each loop again, but considering the current through 50 Ohm resistor as the sum of currents in both loops (be careful of sign).

Reply 2

MonoAno555

Original post by lordaxil

K, spent more time doing this question and want to know if this method works (link has a diagram for more context):

https://postimg.cc/fVgkmWmL

By setting up 2 simultaneous equations:

9v = 300I1 + 50I2 and 6v = 400I1 - 350I2,

I simplify both to get I1 = 69/2500 and I2 = 9/625

Hence as I2 flows through the 50 ohm resistor, Power dissipated = (9/625)² x 50 ≈ 0.0104w (3sf)

Reply 3

lordaxil

Almost there - method is right, but check the currents through 150 and 250 Ohm resistors. I think they should be I₂ - I₁ (rather than
I₁ - I₂).

Reply 4

MonoAno555

Original post by lordaxil

Almost there - method is right, but check the currents through 150 and 250 Ohm resistors. I think they should be I₂ - I₁ (rather than
I₁ - I₂).

oop you are right I think. this time:

by solving 9 = 300I1 + 50I2 and 6 = 450I2 - 400I1 simultaneously, I got I2 = 159/950,

so power dissipated = (159/950)^2 x 50 = 1.40W (3sf)
this right?...

Reply 5

lordaxil

Your equations are correct, but the value of I2 is not. Should be 27/775.