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When two or more electric cells are connected together in a circuit, they can be arranged in either a series or parallel configuration.
A series combination of electric cells is where the cells are connected end-to-end, with the positive terminal of one cell connected to the negative terminal of the next cell, and so on.
The total voltage of the series combination is equal to the sum of the voltages of each individual cell. For example, if two 1.5-volt cells are connected in series, the total voltage of the combination will be 3 volts.
The total current of the series combination is equal to the current flowing through each individual cell, but the internal resistance of each cell is also added together, which can result in a higher overall resistance and reduced current flow.
A parallel combination of electric cells is where the cells are connected side-by-side, with the positive terminals of all the cells connected to one another, and the negative terminals of all the cells connected to one another.
The total voltage of the parallel combination is equal to the voltage of each individual cell. For example, if two 1.5-volt cells are connected in parallel, the total voltage of the combination will still be 1.5 volts.
The total current of the parallel combination is equal to the sum of the currents flowing through each individual cell, but the internal resistance of each cell is effectively reduced, which can result in a lower overall resistance and increased current flow.
Problem (IIT JEE 2011): Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB (in volts) is,
Solution: Let $I$ be the current in the loop (see figure).
The Kirchhoff's loop law, \begin{align} 6-3-2I-1I=0, \end{align} gives $I={1}\;\mathrm{A}$. Now, going from B to A along upper branch, we get, \begin{align} V_B-1+6=V_A, \end{align} which gives \begin{align} V_A-V_B={5}\;\mathrm{V}. \end{align}
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