(a) What is the potential difference Va d in the circuit of Fig. P25.62? (b) What is the termina

(a) What is the potential difference Va d in the circuit of...

Key Concepts

Terminal Voltage

Terminal voltage refers to the voltage measured across the output terminals of a battery when it is connected to an external circuit. It is influenced both by the battery's electromotive force (emf) and any internal resistance which causes a voltage drop when current flows. This concept is crucial for predicting the effective voltage available to power external loads.

Kirchhoff's Loop Rule

Kirchhoff's Loop Rule is a fundamental principle used in circuit analysis that states the sum of the electromotive forces (emfs) and voltage drops around any closed loop in a circuit must equal zero. This concept allows for the setup of equations that account for all energy contributions and drops, making it possible to solve for unknown currents, potential differences, and terminal voltages within complex circuits.

Potential Difference

The potential difference, or voltage, is the work done per unit charge as it moves between two points in an electric field. It represents the energy available to drive the current between those points in a circuit. Understanding potential differences is essential for determining how energy is distributed across various circuit elements.

Internal Resistance

Internal resistance is an inherent property of real batteries, arising from the resistance to the flow of charge within the battery itself. When current flows, this resistance results in a voltage drop inside the battery that reduces the terminal voltage compared to the emf. Recognizing the impact of internal resistance is important for accurately modeling battery behavior in any circuit analysis.

Transcript

00:01 For this problem, all the information we need is contained in the circuit diagram.

00:07 In part a, we are asked to find the potential difference vad.

00:17 And so the first thing to note is to pick a direction for your current instances is the greater of the 2 emf.

00:28 I'm going to assume that the current is flowing.

00:33 This way in the circuit.

00:36 So let's start at point vd and so then as we move along here we encountered this emf of 8 volts and then a drop of 0 .5 oms times whatever the value of the current is and then another drop of 8 oms times whatever the current is and then we're at va but we don't know what the current is so it's obvious from this equation that we need to find the current and the way to do that is to note that the the total potential difference if i go from d and come all the way back is zero.

01:45 So let's start at the 8 volt emf and make our way around the entire circuit.

01:57 So we have 8 volt minus i times 0 .5 oms minus i times 8 oms, minus i times 8 oms, minus i times 6 oms minus i times 0 .5 oms minus 4 volts minus i times 9 oms and we picked up a minus sign for it as emf because the current is flowing from the positive to the negative which is in the opposite sense with what you want emf to send current which goes from the negative to the positive.

03:03 So using this equation we can find i so we have 8 volts minus 4 volts divided by 0 .5 oms plus 8 oms plus 6 oms plus 0 .5 oms plus 9 oms equal to the current and when we plug this into calculator we find that 0 .17 a a appears so now that we have the current we can go back and find vad so vad you to defined as va minus vd.

04:09 So if we subtract vd from this equation up here, then we get vd is equal to 8 volts minus 0 .5 oms times 0 .17 amperors minus 8 oms times 0 .17 amperors, minus 8 oms times 0 .17 ampairs.

04:45 And that gives 6 .55 volts.

04:54 So that's the potential difference between a and d.

05:02 So that solves the first part of the problem.

05:06 Now we have part b, and in part b, it asks, what is the terminal voltage of the 4 -volt battery? that will be b .c because the battery is between the points that are labeled b and c.

05:37 So we'll do the same kind of thing where we let's start at vc.

05:43 So we started vc and then we go through the 4 volt battery and then we actually paper up a positive contribution from the resistor because the point b is at a higher potential than the point c.

06:14 So when we cross that resistor instead of having a voltage drop, we gain voltage because we're going from a lower potential to a higher potential.

06:26 So this is times the current 0 .17 a.

06:38 And that is equal to vb.

06:40 So bb minus vc 2 volts and 4 volts plus which gives us 4 .08 volts so this thing here is vbc and there is a part c to the problem let's do it on the next page in part c of the problem we have that the battery with emf of 10 .3 volts and internal resistance of 0 .5 oms is inserted in the circuit at point d, with its negative terminal connected to the negative terminal of the 8 -volt battery...

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