In the circuit shown in Fig. E25.30, the 16.0-V battery is...

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude and direction); (b) the terminal voltage $V_{ba}$ of the 16.0-V battery; (c) the potential difference $V_{ac}$ of point a with respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

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Transcript

00:02 Alright, so in this problem we have the circuit like this.

00:16 This battery here.

00:19 And this battery is 16 volts.

00:21 This is 8 volts.

00:24 And we have the information about the resistance.

00:29 Let me check.

00:30 1 .6, 1 .4.

00:32 So this is 1 .6, this is 9.

00:35 This is 1 .4.

00:36 And the rest one is 5.

00:41 So in part a, i want to find out the total current in the circuit.

00:44 So, because as you can see that these two batteries, they're connecting series, right? so we have the total emf equal 24 volts.

00:57 They are by the total resistance, which is 1 .6 plus 1 .4 plus 9 plus 5.

01:04 So this gives you 1 .41 ampiers, okay? and direction is pointing to this direction, pointing down, okay? and then part b, we want to find out the potential difference.

01:16 Between vba.

01:18 So this is point a and this is point b.

01:23 So vba, it's simply equal to e, 16 volts, minus i times r.

01:30 I times, so the r i mean, i mean this one, okay? so 1 .6 homes.

01:35 So it's 16 volts minus the current times 1 .6.

01:40 It's a times, okay? so we see that the final result is 13 .7 volts.

01:46 Okay? and part c, want to find out vac.

01:54 So i'm per c...