If cells are joined in parallel, they have _______.
Components connected in parallel provide alternative pathways for current flow. When cells are connected in parallel, the total voltages that they provide does not change. For example, when two or more 2 V cells are connected in parallel, they still provide a total voltage of 2 V.
If the internal resistance of a cell is 20 ohm and resistance of the circuit is also 20 ohm. Will the current remain same whether the given n identical cells are in series or in parallel?
In series connection of cells total internal resistance=nr total resistance of circuit=nR current is same potential difference is different so,I=nv/nR+nr I=V/R+r current is same as before in parallel connection of cells total internal resistance=r/n total resistance of circuit=R/n total current=nI potential difference is same nI=V/r/n+R/n I=V/R+r current is same as before.
Three identical cells, each of 4 V and internal resistance r, are connected in series to a 6 ohm resistor. If the current flowing in the circuit is 2 A. The internal resistance of each cell is
Total emf = 3 x 4 = 12 V
Total resistance = 6 + 3r
Current in the circuit = 2 A
Using ohm's law
2 = 12/ (6+3r)
2(6+3r)=12
12+6r=12
6r = 12-12 = 0
r = 0/6 = 0
When number of identical cells, in parallel combination are increased, the voltage of the circuit will
The voltage developed by the cells in parallel connection cannot be increased by increasing the number of cells present in the circuit. It is because they do not have the same circular path. In parallel connection the connection provides power based on one cell.
The voltage will remain the same.
When numbers of cells in series combination are increased, the current of the circuit will
When the battery (no of cells together) is connected then potential remains same. So increase in no of cell will increase the current.
If there are n cells of emf E_{1}, E_{2}, …E_{n} and of internal resistances r_{1}, r_{2} …r_{n} , connected in parallel, the combination is equivalent to a cell of emf Eeq such that
The equivalent emf of two cells (with ε_{1} > ε_{2}) in series if we connect two negative electrodes of adjacent cells is
When we connect cells to form a battery, we join opposite ends(+ve with -ve and vice versa) to get supporting emf, i.e. net emf get added, but when we join the same polarities(-ve with -ve, like in the question), they oppose each other, hence net emf is the difference b/w them.
The expression for equivalent emf when two cells of emf’s
ε_{eq} = ε_{1} - ε_{2} (ε_{1}> ε_{2})
The internal resistance of two identical cells, with same current whether joined in series or parallel to a 1Ω resistor is
1^{st} case[series]
I=E+E/r+r+1-----1
2^{nd} case[parallel]
I=E[E will be same in parallel]/(1/r)+(1/r)+(1/1)----2
Current in both cases are the same.
1=2
2E/2r+1=E/(2/r)+1
R=1 Ω
If cells are joined in series, they have ________.
In series, cells are joined end to end so that the same current flows through each cell. In case if the cells are connected in series the emf of the battery is connected to the sum of the emf of the individual cells. Suppose we have multiple cells and they are arranged in such a way that the positive terminal of one cell is connected to the negative terminal of the other and then again the negative terminal is connected to the positive terminal and so on, then we can that the cell is connected in series.
Three identical cells, each of 2V and internal resistance of 0.2 ohm are connected in series to an external resistance of 7.4 ohm. The current in the circuit will be
From the Question,
Emf (E) = 2 V
Internal Resistance (r) = 0.2 Ω
Resistance (R) = 7.4 Ω
Three cells of 2V and 0.2 each are connected in series.
Consider two cells connected to each other.
The total potential difference would be :
Total Potential Difference in a cell is given as :
V=E-IR
Thus,
⇒E-Ir=E1-Ir_{1}+E_{2}+Ir_{2}
⇒E_{eq}=E_{1}+E_{2} and r_{eq}=r_{1}+r_{2}
Therefore,
For three cells each of emf 2V,
E=2+2+2=6V
For three cells with internal resistance 0.2 Ω,
R=.02+.02+.02=0.6 Ω
We know that,
I=E/r+R
Implies,
I=6/0.6+7.4
I=6/8
I=0.75A
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