Determine Av=Vo / Vs for the network of Fig. 181 if the source has an internal resistance of 1.2

Determine Av=Vo / Vs for the network of Fig. 181 if the...

Key Concepts

Internal Resistance

Internal resistance refers to the inherent resistance present within a voltage source. This non-ideal characteristic of real-world sources causes a drop in output voltage when current flows, and it must be taken into account in circuit analysis, particularly when calculating voltage drops, as it directly affects the performance of the overall circuit.

Load Resistance

Load resistance is the resistance of the component, device, or system that draws power from the source. In circuit analysis, its value is critical because it influences the final voltage distribution and current flow. A load connected to a voltage divider can alter the effective resistance seen by the source, affecting the voltage division ratio and overall circuit performance.

Voltage Divider

The voltage divider is a fundamental circuit concept where the output voltage across one of a pair (or more) of series-connected resistors is a fraction of the input voltage. It relies on the principle that the voltage is divided among components in proportion to their resistances, allowing one to determine how much of the input voltage appears across a specific resistor or load in the circuit.

Thevenin Equivalent Circuits

The concept of Thevenin equivalent circuits simplifies the analysis of complex circuits by representing them as a single voltage source in series with a single resistor. This technique is especially useful when analyzing how varying load resistances affect the voltage and current in a circuit, enabling easier calculations and understanding of system behavior under different loading conditions.

Transcript

00:01 So this question belongs to the current electricity in which we have to determine the current i drawn from the battery v equals to 12 volt supply with internal resistance small r equals to 0 .5 oom by the infinite resistance and each register is 1 oom.

00:15 So the given circuit diagram is this.

00:18 Now we have to calculate the equivalent resistance of this circuit across these two points.

00:23 So let us consider this equivalent resistance of the circuit is r.

00:30 So, this is the infinite resistance.

00:32 So the equivalent resistance is r.

00:33 So we can reduce, remove this terms which is infinite terms.

00:38 So we will get the resistance for these terms will also be r because this is infinite circuit.

00:44 So we will reduce to this circuit.

00:46 This will be the reduced circuit for this question.

00:50 This is resistance r which is equivalent and this is one ome and this is one ome and this is again one ome and across these two points the equivalent resistance.

01:00 Is again capital r.

01:02 So now for this circuit we can write that do these two resistances are in parallel.

01:08 So the equivalent of these two resistances will be r multiplied by 1 divided by r plus 1.

01:14 So it will be equals to r divided by r plus 1.

01:17 Now this circuit will reduce to this.

01:20 So this will be the reduction of this circuit and this will be the battery here...

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