The V = 10.50 V battery in the figure below is removed from the circuit and rein

Question

The V = 10.50 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 11.00 ?.) (a) Find the current in each branch. upper branch, middle branch, lower branch (b) Find the potential difference Vab of point a relative to point b. 12. 24 points | Previous Answers YF14 26.P029 My Notes Ask Your The V- 10.50 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R-11.00.) (a) Find the current in each branch. upper branch middle branch 1.45 lower branch 0.204 (b) Find the potential difference V 10.5 of point a relative to point b.

Explanation / Answer

Apply KVL to the upper loop

-3 i1 - 4 i2 +5 V - i2- 2 i1+10.5 = 0

-5 i1- 5 i2+ 15.5 = 0

in lower loop

-11 i1( i1-i2)+ i2- 5+ 4 i2 = 0

-11 i1+ 16 i2- 5 = 0

i2 = 11 i1+ 5/16

substitue above equation to first equation

-5 i1- 5 (11 i1+ 5/16)+ 15.5 = 0

-5 i1 - 55 i1- 25/16 - 15.5 = 0

-80 i1 - 55 i1- 25 - 248 = 0

i1 = 2.022 A

i2 = 1.70 A

lower branch i1-i2 = 0.322 A

(b)

Va- Vb = 3 i1 + 4 i2

= 3(2.022) + 4( 1.7)

=12.866 V

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