Current Electricity Concept Page - 8

Definition

Internal resistance

The resistance offered by the electrolyte inside the cell to the flow of current is called the internal resistance of the cell (r).
r=ϵImax$r={\displaystyle \frac{ϵ}{I_{max}}}$ where Imax$I_{max}$ is the maximum current that can be drawn from the cell.

Definition

Maximum current drawn from a cell

The maximum current that can be drawn from a cell is for R=0$R=0$ and it is
Imax=Er$I_{max}={\displaystyle \frac{E}{r}}$ where r$r$ is the internal resistance of the cell.

Definition

Circuit with a singe cell and different resistor combinations

Current will flow from a battery of 2 V$2V$, 4$4$ ohm resistor, and the three 4$4$ ohm resistors in parallel, and back into the battery.Find current in the battery.

Solution:
 24+4/3=2×34×4=38A${\displaystyle \frac{2}{4+4/3}}={\displaystyle \frac{2\times 3}{4\times 4}}={\displaystyle \frac{3}{8}}A$Current in the ammeter = 13×38=18A${\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{8}}={\displaystyle \frac{1}{8}}A$

Definition

Equivalent emf/resistance of cells in parallel

If emf of each cell is identical, then the emf of the battery combined by n numbers of cells connected in parallel, is equal to the emf of each cell. The resultant internal resistance of the combination is,
r=(1r1+1r2+1r3+........+1rn)−1$r={({\displaystyle \frac{1}{r_1}}+{\displaystyle \frac{1}{r_2}}+{\displaystyle \frac{1}{r_3}}+........+{\displaystyle \frac{1}{r_n}})}^{-1}$

Definition

Equivalent emf/resistance of cells in series and parallel

Again assume emf of each cell is E$E$ and internal resistance of each cell is r$r$. As n$n$ numbers of cells are connected in each series, the emf of each series as well as the battery will be nE$nE$. The equivalent resistance of the series is n$n$r. As, m$m$ number of series connected in parallel equivalent internal resistance of that series and parallel battery is nr/m$nr/m$.

Definition

Equivalent emf/resistance of cells in series

If E$E$ is the overall emf of the battery combined by n$n$ number cells and E1,E2,E3,En$E_1,E_2,E_3,E_n$ are the emfs of individual cells.
Then E=E1+E2+E3+...+En$E=E_1+E_2+E_3+...+E_n$
Similarly, if r1,r2,r3,rn$r_1,r_2,r_3,r_n$ are the internal resistances of individual cells, then the internal resistance of the battery will be equal to the sum of the internal resistance of the individual cells i.e.
r=r1+r2+r3++rn$r=r_1+r_2+r_3++r_n$

Example

Electrical circuits consisting of resistors and cells

In the circuit shown in the figure, the ammeter reading can be given by:
Let, IA$I_A$ is current through ammeter, I4$I_4$ is current through 4$4$ ohm resistor and I12$I_{12}$ is current through 12$12$ ohm resistor.
 I4=64+0.6=1.30A$I_4={\displaystyle \frac{6}{4+0.6}}=1.30A$
 I12=612+0.6=0.47A$I_{12}={\displaystyle \frac{6}{12+0.6}}=0.47A$
Hence,
IA=I4−I12$I_A=I_4-I_{12}$
IA=1.30−0.47=0.83A$I_A=1.30-0.47=0.83A$

Law

The junction law

The sum of all the currents directed towards a point in a circuit is equal to the sum of all the currents directed away from the point.

Law

The loop law

The algebraic sum of all the potential differences along a closed loop in a circuit is zero.

Example

Equivalent resistance using Kirchoff's laws

⇒6I+3(I+I2)=E$\Rightarrow 6I+3(I+I_2)=E$.................1
3(3−I)+6(3−(I+I2))=E$3(3-I)+6(3-(I+I_2))=E$
9I+3I2=E$9I+3I_2=E$
27−9I−6I2=E$27-9I-6I_2=E$
Applying KVL in loop 1,
6I−3(3−I)=0$6I-3(3-I)=0$
2I=3−I$2I=3-I$
3I=3⇒I=1$3I=3\Rightarrow I=1$       ...........................  2

Using 2 , 
9I+3I2$9I+3I_2$
=27−9I−6I2$=27-9I-6I_2$
9+3I2=18−6I2$9+3I_2=18-6I_2$
9I2=9$9I_2=9$
I2=1A$I_2=1A$
From 1, E=12$E=12$R$R$=E3${\displaystyle \frac{E}{3}}$
R$R$=4Ω

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