The potential difference between the terminals of a cell is found to be 3 volts

In a serially connected circuit the voltage across is directly proportional to the resistance in the circuit. That is, V ∝ R. So, V t o t a l ​ V ​ = R t o t a l ​ R ​ The total resistance R in the circuit is R = 2 + 3 + 5 = 1 0 ohm. Since, the resistance of the 3 ohms resistor is only three tenth of the total resistance, the applied voltage is given as 1 0 3 ​ × 2 = 0 . 6 volt Hence, t he potential difference between the terminals of 3 ohm resistance will be 0.6 volt.

(i) EMF of the cell is equal to maximum potential difference across the two electrodes of cell corresponding to zero current. Thus emf of the cell, `epsilon = 1.4 V`. (ii) Max. current is drawn from the cell when the terminal pot. Diff. is zero. Therefore `I_(max) = 0.28 A` (iii) Internal resistance, `r=epsilon/I_(max) = (1.4 V)/(0.28A)= 5 Omega`

Electrical sources and internal resistance An electrical cell is made from materials (metal or chemicals, for example). All materials have some resistance. Therefore, a cell must have resistance . This resistance is called the internal resistance of the cell. A cell can be thought of as a source of electromotive force (EMF) with a resistor connected in series . When a load resistance is connected, current flows through the cell and a voltage develops across the internal resistance. This voltage is not available to the circuit so it is called the lost volts, \(V_\) This equation can be written in different forms, eg \(E= I(R + r)\) . To solve problems on internal resistance it should be remembered that such circuits involve using a series circuit with the internal resistance and the load.

Potential differences across the terminals of a cell were measured (in volt) against different currents (in ampere) flowing through the cell. A graph was drawn which was a straight line ABC as shown in figure Determine from graph (i) emf of the cell (ii) maximum current obtained from the cell and (iii) internal resistance of the cell. (i) EMF of the cell is equal to maximum potential difference across the two electrodes of cell corresponding to zero current. Thus emf of the cell, ε = 1.4 V. (ii) Max. current is drawn from the cell when the terminal pot. Diff. is zero. Therefore I max = 0.28 A (iii) Internal resistance, r = ε I max = 1.4 V 0.28 A = 5 Ω एक सेल से प्रवाहित धारा (ऐम्पियर में ) तथा उसके लिए सेल के सिरों का विभवांतर (वोल्ट में ) मापा गया है । प्राप्त प्रेक्षणों से ग्राफ खींचने पर चित्र की भांति एक सरल रेखा ABC प्राप्त हुई है । ग्राफ से ज्ञात कीजिये - (i) सेल का वि वा बल (emf ) तथा सेल से उपलब्ध अधिकतम धारा । (ii ) सेल का आंतरिक प्रतिरोध ।

\( \newcommand\) • • • • • • • • • • • • Learning Objectives • To understand the basics of voltaic cells • To connect voltage from a voltaic cell to underlying redox chemistry In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements. Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows: \[\ce\): Electrochemical Cells. A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necess...

In this explainer, we will learn how to calculate the potential difference provided by a cell based on the amount of work it does to separate charge. Recall the properties of charge. An object or particle can be either positively charged, negatively charged, or neutral. Charge is measured in units of coulombs, C. Like charges repel each other, and opposite charges attract each other. This means that a positive charge and a negative charge will experience an attractive force between them. The attractive force pulls them toward each other. If we wish to separate two opposite charges, we must overcome the attractive force between them. This means we must do work on the charges. In order to increase the distance between the charges, we must transfer energy to the charges, so they are able to overcome the attractive force between them. When we separate the charges, we create an electric potential difference between them. The electric potential difference tells us how much work has been done on the charges in order to separate them. Complete the following sentence: The separation of the positive and negative charges creates an between them. • electric potential difference • electric current Answer The answer is A, electric potential difference. The picture shows a positive charge and a negative charge, with a distance in between them. Hence, the picture shows a pair of opposite charges, which have been separated. To separate opposite charges, work must be done to overcome the at...