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Question: How many cells each marked \(6V - 12A\) should be connected in mixed grouping so that he marked \(24...

How many cells each marked 6V12A6V - 12A should be connected in mixed grouping so that he marked 24V24A24V - 24A
A.A. 44
B.B. 88
C.C. 1212
D.D. 66

Explanation

Solution

In this problem the cell which is having emfemf of 6V6V and current of 12A12A is connected in different combinations that is series and parallel combination to obtain the emfemf of 24V24V and current of 24A24A .

Complete step-by-step solution:
If the cells are connected in series the resultant emfemf of the battery will be equal to sum of the emfemf of the individual cell and the current remains same .If EE is the resultant emfemf of the of the battery with nn number of cells and E1,E2,E3,E4.......En.{E_1},{E_2},{E_3},{E_4}.......{E_n}. are the emfemf of individual cells. Then
E=E1+E2+E3+E4.......+En.E = {E_1} + {E_2} + {E_3} + {E_4}....... + {E_n}.……… (1)\left( 1 \right)
II remains the same.
If the cells are connected in parallel the resultant current of the battery will be equal to sum of the current of the individual cell and the emfemf remains same .If II is the resultant current of the of the battery with nn number of cells and I1,I2,I3,I4.......In.{I_1},{I_2},{I_3},{I_4}.......{I_n}. are the current of individual cells. Then
I=I1,I2,I3,I4.......In.I = {I_1},{I_2},{I_3},{I_4}.......{I_n}. ……… (2)\left( 2 \right)
EE remains the same.

Given:
\Rightarrow Each cell have emfemf of 6V6V and current of 12A12A
ε=6V,I=12A\varepsilon = 6V,I = 12A

This cell should be connected to get resultant emfemf of 24V24V and current of 24A24A
\Rightarrow When 44 cells are connected in series the resultant emfemf is 24V24V and the current remains the same that is 12A12A because the current is same in series combination
ε=24V,I=12A\varepsilon = 24V,I = 12A

\Rightarrow Then this combination is connected in parallel the resultant emfemf remains the same that is 24V24V because voltage or emfemf remains same in parallel combination and the current will be 24A24A
ε=24V,I=24A\varepsilon = 24V,I = 24A

Therefore, 88 cells are connected in mixed group as shown above to get, ε=24V,I=24A\varepsilon = 24V,I = 24A
Hence, option BB is correct. That is 88

Note: In series connection the current will be the same and potential difference or voltage or emfemf will be different. Whereas in parallel combination the current will be different and the potential difference will be the same.