I'm not sure how to do question 9.5a and 9.5b. I read the direction for 9.5a but i'm still unsure. it says when you solve for current, current is equal to 1. How did they solve for it?
Could somebody explain these 2 to me?
Thanks.
I'm assuming this is 9.7a and 9.7b in my book, but if I'm wrong, please let me know.
9.7a
The second part of that question (determining I
3) is tricky at first, but it makes it easy once you see it. They tell you that I
1 = 2 amps. The current splits at the junction such that I
1 = I
2 + I
3. If I
1 is 2 amps, then I
3 must be less than 2 amps. Only choice A has I
3 less than 2 amps. You gotta love mulitple choice!
😀
To solve for 1 amp from the information given (they tell you that R
2 = R
3), you need to notice that I
2 = I
3 when the resistors have the same current. I
2 + I
3 = 2 amps and I
2 = I
3, so I
2 = 1 amp and I
3 = 1 amp.
9.7b
The two parallel pathways have resistances of 6 ohms through the upper path (R
1 = 3 ohms and R
2 = 3 ohms, so R
1 + R
2 = 6 ohms) and 3 ohms through the lower path. The equivalent resistance for the circuit is 2 ohms (1/R
eq = 1/3 + 1/6 = 1/2, so R
eq = 2 ohms). This means that that the battery voltage must be 12 V (6 A x 2 ohms = 12 V).
Because R
1 = R
2, I
1 = I
2.
V
total = I
1R
1 + I
2R
2 where I
1R
1 = I
2R
2.
12 = I
1R
1 + I
2R
2, so I
2R
2 = 6V (choice C).
For current, we know that parallel pathways share the same voltage drops, so I
1R
1 + I
2R
2 = I
3R
3 = 12 V. You can either look at the junction rule to see that I
total = 6 amps and I
3 = 2 x I
1, so I
1 = 2 amps and I
3 = 4 amps.
You could have also gotten in by knowing V
3 = 12 V and R
3 = 3 ohms, so I
3 = 4 amps.
Hope this makes sense.
