mbrmbrg
- 485
- 2
1. the problem
In Figure 27-69 (see attached), the ideal each resistance is 4.50 and V1 = 11.0 V.
(a) What is the size and direction of current i1? (Take upward to be positive.)
(No Response)[-2.44] A
(b) What is the size and direction of current i2? (Take upward to be positive.)
(No Response)[-1.33] A
(c) At what rate is energy being transferred at the 4.00 V battery and is the battery supplying or absorbing energy?
(No Response)[5.33] W (No Response)[ supplying ]
(d) At what rate is energy being transferred at the 11.0 V battery and is the battery supplying or absorbing energy?
(No Response)[41.6] W (No Response)[ supplying ]
Attempted Solution
Help needed with part b.
for part a: we ignored the right hand part of the cirucuit, and just did the left-most loop. Then we summed the voltages and set them to zero, and got the correct answer.
for part b: We looked at the two right loops and combined them by saying that there are two resistors in parallel which can be replaced by R_{eq}=\frac{1}{R}+\frac{1}{R})^{-1}. Then we tried ignoring the left loop and summing the voltages in the middle loop (with the top right resistor replaced by R_eq) and setting everything equal to zero. That's the wrong answer.
Frankly, I don't understand why i_2\neq i_1.
And the bloody final is tomorrow
In Figure 27-69 (see attached), the ideal each resistance is 4.50 and V1 = 11.0 V.
(a) What is the size and direction of current i1? (Take upward to be positive.)
(No Response)[-2.44] A
(b) What is the size and direction of current i2? (Take upward to be positive.)
(No Response)[-1.33] A
(c) At what rate is energy being transferred at the 4.00 V battery and is the battery supplying or absorbing energy?
(No Response)[5.33] W (No Response)[ supplying ]
(d) At what rate is energy being transferred at the 11.0 V battery and is the battery supplying or absorbing energy?
(No Response)[41.6] W (No Response)[ supplying ]
Attempted Solution
Help needed with part b.
for part a: we ignored the right hand part of the cirucuit, and just did the left-most loop. Then we summed the voltages and set them to zero, and got the correct answer.
for part b: We looked at the two right loops and combined them by saying that there are two resistors in parallel which can be replaced by R_{eq}=\frac{1}{R}+\frac{1}{R})^{-1}. Then we tried ignoring the left loop and summing the voltages in the middle loop (with the top right resistor replaced by R_eq) and setting everything equal to zero. That's the wrong answer.
Frankly, I don't understand why i_2\neq i_1.
And the bloody final is tomorrow
