The circuit shown in Fig. 𝐄 25.30 contain two batteries, each with an emf and an internal resist

The circuit shown in Fig. 𝐄 25.30 contain two batteries,...

Key Concepts

Series Circuit Analysis

In a circuit where components such as batteries and resistors are connected in series, the same current flows through each component. This allows for the straightforward application of Ohm’s Law across each element and the combined total resistance to determine the circuit current. Series analysis is essential for solving for currents and voltage drops in circuits with multiple series-connected components.

Terminal Voltage

The terminal voltage of a battery is the voltage measured across its terminals when a current is drawn and is given by subtracting the voltage drop across the internal resistance (I × r) from the emf. This concept is crucial in circuits where the battery is under load, as it directly affects the performance and behavior of the circuit elements connected to it.

Potential Rises and Drops (Voltage Profiling)

Identifying and graphing the potential rises and drops across circuit elements is a graphical method to visualize energy changes within a circuit. It involves mapping the increases in potential due to emfs and the decreases due to resistive elements (both external and internal), providing intuitive insight into how the overall voltage is distributed around the circuit loop. This technique is particularly useful in understanding and verifying the energy conservation embodied in KVL.

Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law states that the sum of all electromotive forces (emfs) and voltage drops around any closed loop in a circuit must equal zero. This principle is fundamental in circuit analysis since it allows for the writing of equations that relate the emfs, internal resistances, and other resistive elements in the circuit, enabling the determination of unknown currents and potentials.

Battery Internal Resistance

Batteries are not ideal voltage sources; they possess an internal resistance that causes a voltage drop when current flows. This concept is important because the actual terminal voltage available to the external circuit is reduced by the voltage drop across the battery’s internal resistance. Understanding this difference between the emf and the terminal voltage is key to accurately analyzing circuits involving real batteries.

Transcript

00:01 Hi, in the given problem here, this is a circuit having a resistance, internal resistance of the cell whose emf is 16 .0 volt.

00:21 Its internal resistance is 1 .6 foam.

00:24 This is the terminal b and this is terminal a.

00:30 Then an external resistance attached.

00:34 With it having a value of 5 .0 oom, then another cell having internal resistance as 1 .4 om and having the polarities like this, having emf 8 .0 volt and the terminal c and finally one more external resistance, 9 .0.

01:10 So as both the source, of emf are sending the currents in opposite direction.

01:19 So the net emf will be in first part of the problem, the net emf of the circuit will be 16 .0 minus 8 .0 volt, which comes out to be 8 .0 volt and sending the current in a counterclockwise direction.

01:52 From the figure.

01:55 The net resistance of the circuit as all the resistors are joined in series.

02:01 So this is 1 .6 plus 5 .0 plus 1 .4 and plus 9 .0 .m which comes out to be 17 .0.

02:21 Current in the circuit will be given by om's law as net emf e by net resistance r means this is 8 .0 divided by 17 .0.

02:38 So finally this current in the circuit comes out to be 0 .47 ampere.

02:47 And here it becomes the answer for the first part of this problem.

02:54 In the second part of the problem we have to find terminal potential difference across the first cell means across a and b.

03:04 So we know the terminal potential difference is having an expression as the emfe minus the current passing through the cell into its internal resistance.

03:18 So for emf of this cell, this is 16 .0 volt minus current as current remains same in the series so this is 0 .47 into internal resistance of this upper cell which is 1 .6 so finally this terminal potential difference we ab here comes out to be 16 .0 minus 0 .753 volt or we can say this is 15 .25 5 volt and it becomes the answer for the second part of this problem.

04:08 Then in the third part of the problem, we have to find the potential difference between terminal a and terminal c.

04:19 Means we have to find va minus vc.

04:23 To find it, we start measuring the potentials from point a then the potential is dropping across this resistance across this resistance and as the terminals of this battery are joint opposite so that 8 volt you saw will also be considered as a drop in the potential and then ultimately all these algebraic sons will be equal to bc so we can write it like v a minus 5 .0 into i current in the suns current in the size circuit which is 0 .4 7, then minus 1 .4 into again the same current 0 .47, then 8 volt of this lower battery should be equal to vc.

05:15 So finally rearranging the terms we get vm minus vc is equal to 5 plus 1 .4.

05:24 Out of these two we can take this 0 .47 as a common out and plus 8 .0 volt.

05:31 So finally this va minus vc comes out to be 11 .0 volt, which becomes the answer for the third part of the problem.

05:41 Finally, in the last part of the problem, we have to plot...

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