# Derive an expression for the current in a circuit with external resistance R when

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Derive an expression for the current in a circuit with external resistance R when (a) n identical cells of emf E and internal resistance r are connected in series (b) m identical cells are connected in parallel
asked Jul 25, 2014 in Physics

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(A) Series combination

• Figure below shows the two cells of emf's E1 and E2 and internal resistance r1 and r2 respectively connected in series combination through external resistance

• Points A and B in the circuit acts as two terminals of the combination
• Applying kirchoff's loop rule to above closed circuit
-Ir2-Ir1-IR+E1+E2=0
or
I=E1+E2/R+(r1+r2)
Where I is the current flowing through the external resistance R
• Let total internal resistance of the combination by r=r1+r2 and also let E=E1+E2 is the total EMF of the two cells
• Thus this combination of two cells acts as a cell of emf E=E1+E2 having total internal resistance r=r1+r2 as shown above in the figure

(B) Parallel combinations of cells

• Figure below shows the two cells of emf E1 and E2 and internal resistance r1 and r2 respectively connected in parallel combination through external resistance

• Applying kirchoff's loop rule in loop containing E1 ,r1 and R,we find
E1-IR-I1r1=0 ------------------------(1)
Similarly applying kirchoff's loop rule in loop containing E2 ,r2 and R,we find
E2-IR-(I-I1)r2=0 ------------------------(2)
• Now we have to solve equation 1 and 2 for the value of I,So multiplying 1 by r2 and 2 by r1 and then adding these equations results in following equation
IR(r1+r2)+r2r1I-E1r2-E2r1=0
which gives

We can rewrite this as

E is the resulting EMF due to parallel combination of cells and r is resulting internal resistance.
answered Jul 28, 2014 by (4,920 points)