Name: Kirchhoff's Second Law
Uploaded: 2019-07-19
Duration: 5 min 21 s
Description: Kirchhoff’s first law is very handy when calculating currents. Here, we explore his second potent law and use it to reduce complicated loops. Watch and learn.

Question

Obtain the expression for the effective emf and the effective internal resistance of two cells connected in parallel such that the currents are flowing in the same direction.

Accepted Answer

Let $E_{1}$ and $E_{2}$ be the emfs of two cells connected in parallel. Let $r_{1}$ and $r_{2}$ be their internal resistance respectively.

$I_{1}$ and $I_{2}$ are the currents leaving the positive electrodes of the cells.

According to diagram $I = I_{1} + I_{2}$

Let $V_{1}$ and $V_{2}$ be the potentials at $B_{1}$ and $B_{2}$ respectively.

Potential difference across the first cell is $V_{1} - V_{2} = E_{1} - I_{1} r_{1}$

Potential difference across the second cell is also $V_{1} - V_{2} = E_{2} - I_{2} r_{2}$

Hence,

$I_{1} = \frac{E_{1} - V}{r_{1}}$ and $I_{2} = \frac{E_{2} - V}{r_{2}}$

$I = I_{1} + I_{2} = \frac{E_{1} - V}{r_{1}} + \frac{E_{2} - V}{r_{2}}$

$I = I_{1} + I_{2} = \frac{E_{1}}{r_{1}} - \frac{V}{r_{1}} + \frac{E_{2}}{r_{2}} - \frac{V}{r_{2}}$

$I = I_{1} + I_{2} = \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} - V (\frac{1}{r_{2}} + \frac{1}{r_{1}})$

$I = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}} - V (\frac{r_{2} + r_{2}}{r_{2} r_{1}})$

$V (\frac{r_{2} + r_{2}}{r_{2} r_{1}}) = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}} - I$

$V (r_{2} + r_{2}) = E_{1} r_{2} + E_{2} r_{1} - r_{1} r_{2} I$

$V = \frac{E_{1} r_{2} + E_{2} r_{1}}{(r_{2} + r_{2})} - \frac{r_{1} r_{2} I}{(r_{2} + r_{2})}$

If the parallel combination of cells is replaced by a single cell between $B_{1}$ and $B_{2}$ of emf $ε_{P}$ and internal resistance $R_{P}$ then, $V = ε_{P} - I R_{P}$

Comparing the above equations

$ε_{P} = \frac{E_{1} r_{2} + E_{2} r_{1}}{(r_{2} + r_{2})}$ and $R_{P} = \frac{r_{1} r_{2}}{(r_{2} + r_{2})}$

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