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

Two cells of emf

E_{1}

and

E_{2}

, internal resistance

r_{1}

and

r_{2}

are connected in parallel such that they send current in same direction. Derive an expression for equivalent resistance and equivalent emf of the combination.

Accepted Answer

Consider two cells having electromotive force of $E_{1}$ and $E_{2}$ , and internal resistances $r_{1}$ and $r_{2}$ respectively.

Assuming the cells are connected parallelly:

If current $I_{1}$ and $I_{2}$ are the current through those respective cell then, $I = I_{1} + I_{2}$ ......................... $(1)$

Suppose, $V$ be the potential difference across each cell.

Then, considering first cell, $V = E_{1} - I_{1} r_{1}$ ......................... $(2)$

So, $I_{1} = \frac{E_{1} - V}{r_{1}}$ ......................... $(3)$

Now, considering second cell, $V = E_{2} - I_{2} r_{2}$ ......................... $(4)$

So, $I_{2} = \frac{E_{2} - V}{r_{2}}$ ......................... $(5)$

Substituting, equation $(3)$ and $(5)$ in $(1)$ , we get, $I = \frac{E_{1} - V}{r_{1}} + \frac{E_{2} - V}{r_{2}}$

$I = \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} - V (\frac{1}{r_{1}} + \frac{1}{r_{2}})$ ......................... $(6)$

Now, assuming a single cell is equivalent to those two cells, its emf is $E_{e q}$ and internal resistance $r_{e q}$ .

Then, $I = \frac{E_{e q}}{r_{e q}} - \frac{V}{r_{e q}}$ ......................... $(7)$

Now, by comparing equation $(6)$ and $(7)$ , we get:

$\frac{E_{e q}}{r_{e q}} = \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}$ ......................... $(8)$ and $\frac{1}{r_{e q}} = (\frac{1}{r_{1}} + \frac{1}{r_{2}}) \Rightarrow r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$ ......................... $(9)$

Substituting equation $(9)$ in $(8)$ , we get: $E_{e q} = (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) r_{e q} \Rightarrow E_{e q} = (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) \frac{r_{1} r_{2}}{r_{1} + r_{2}} \Rightarrow E_{e q} = \frac{r_{2} E_{1} + r_{1} E_{2}}{r_{1} + r_{2}}$

Hence, the expression for equivalent resistance and equivalent emf of the combination is $E_{e q} = \frac{r_{2} E_{1} + r_{1} E_{2}}{r_{1} + r_{2}}$ .

Two cells of emf $E_{1}$ and $E_{2}$ , internal resistance $r_{1}$ and $r_{2}$ are connected in parallel such that they send current in same direction. Derive an expression for equivalent resistance and equivalent emf of the combination.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

Two cells of emf E1 and E2, internal resistance r1 and r2 are connected in parallel such that they send current in same direction. Derive an expression for equivalent resistance and equivalent emf of the combination.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

Two cells of emf $E_{1}$ and $E_{2}$ , internal resistance $r_{1}$ and $r_{2}$ are connected in parallel such that they send current in same direction. Derive an expression for equivalent resistance and equivalent emf of the combination.