Question

 The 5.00 V battery in (Figure 1) is removed from the circuit and replaced by a 15.00 V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure.

 Part A

 Find the current in the upper branch.

 Part B

 Find the current in the middle branch.

 Part C

 Find the current in the lower branch.

 Part D

 Find the potential difference Vab of point a relative to point b.

Answer 1

Answer 2

For Mastering Physics:

a) 0A

b) 1.00A

c) -1.00A

d) 4V

I got all of them wrong, and the hw gave it to me

Answer 3

I tried my best.

First, calculate the current of the middle-branch and bottom-branch.

In the circuit (1ohm-15V-4ohm-10ohm), I_mid = V/R = 15/(1+4+10) = 1A

for the middle branch: 1A to the left (positive)

for the bottom branch: 1A to the right (negative)

Because I_up+I_mid = I_bottom, I_up = 0

Ans:

(a)0A

(b)+1A

(c)-1A

(d) *may not be correct*

point a: neutral (I = 0)

point b->a: 4ohm * 1A - 0 = 4V