(a) Find the potential of point a with respect to point b in Fig. P 26.67 . (b) If points a and

(a) Find the potential of point a with respect to point b in...

Key Concepts

Kirchhoff’s Circuit Laws

Kirchhoff’s laws, including the voltage law (the sum of potential differences around any closed loop must equal zero) and the current law (the sum of currents entering a junction equals the sum leaving it), are vital in analyzing complex circuits. They help to determine unknown voltages and currents by applying the principles of energy conservation and charge conservation.

Ohm’s Law and Current Flow

Ohm’s Law relates the potential difference across an element to the current flowing through it and the element’s resistance (V = IR). This relationship is essential for determining the current when the voltage and resistance in a circuit path are known, particularly in the analysis of circuits containing batteries and resistive elements.

Battery Electromotive Force (emf)

A battery is a device that provides a constant potential difference (emf) between its terminals, which drives current through the circuit. Understanding a battery’s role in establishing potential differences and influencing current flow is key in circuit analysis.

Potential Difference

Also known as voltage, the potential difference between two points is a measure of the work required per unit charge to move a charge between those points. It is the fundamental driving force for current in a circuit and is calculated from the difference in electric potentials.

Electric Potential

This concept refers to the energy per unit charge at a point in an electric field, determined by the work done to bring a charge from a reference point (usually taken as zero potential) to that point. It is a scalar quantity that helps analyze energy changes in circuits and electrostatic systems.

Equipotential Connections

In circuits, conductors with negligible resistance can be treated as equipotential, meaning that every point along the conductor has the same electric potential. This concept is crucial for understanding how connections (like wires with negligible resistance) influence the distribution of voltage and current in a circuit.

Transcript

00:01 For this problem on the topic of direct current circuits, we are given a diagram of a circuit and are asked to find the potential of point a with respect to b as it is in the diagram.

00:11 And then we are to assume that a and b are connected by a wire with negligible resistance and asked to find the current in the need battery.

00:20 So firstly, if we look at the circuit for part a, the break in the circuit between a and b mean there's no there's no current in the middle branch that has the 3 -oom resistor and each time.

00:31 10 -volt battery.

00:33 And so the circuit therefore has a single current path, which in the diagram we can see is the outer loop.

00:42 So we need to find the current and we then we can calculate the potential drops across the resistors.

00:48 So we'll calculate vab by traveling from a to b and keeping track of the potential changes along the path that we take.

00:57 So we'll apply the loop rule to the outer loop and we get the following.

01:03 Plus 12 volts.

01:05 Minus the current i that flows in this outer loop into one om plus two oms plus two oms plus one ome minus eight volts minus i into two oms plus one om and according to the loop rule this must equal to zero so from here we can simply rearrange and find the one unknown which is the current in this outer loop.

01:57 So the current i is equal to 12 volts minus 8 volts over 9 oms.

02:11 This gives us a current of 0 .444amphs.

02:19 Now that we have the current to find the potential difference vab will start at a we'll start at b rather and travel to a and we'll add up the potential rises and drops along the path now again the one oom and three oom resistance and resistors in the middle branch have no current through them and hence no voltage so again if we go from b to a we have vb minus 10 volts plus 12 volts minus the current i which you calculated into one oom plus 1 om plus 2 oms is equal to va.

03:19 So from here we can see that vab, which is simply va minus vb, so you rearrange the equation above, is equal to 2 volts minus the current i, which we calculated to be 0 .444 -ampiers times the resistance of 4 -on -oom.

03:51 Gives us the potential difference across ab or the potential from a to b to be positive 0 .22 volts.

04:08 So we have the potential vab and this tells us that a is at a higher potential since this value is positive.

04:19 So that's our answer for part a.

04:25 Now for part b we assume that a and b is connected by a wire with negligible resistance.

04:34 And with a and b connected by a wire, there are now three current branches, as we can see in our diagram...

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