Question

Six lead accumulators, each of emf 2.0 V and the internal resistance 0.015 ohm joined in series to supply current to an external resistance of 8.5 ohm. Find the current drawn from the supply and its terminal voltage

Answer 1

Given,

The emf of one cell = 2.0 V

The total emf of the series combination 6 cells =  6 * 2.0 = 12.0 V

And the internal resistance on each cell = 0.015 Ω

thus the total internal resistance = r =  6* 0.015 = 0.09  Ω

Here , R = 8.5  Ω

So, the current drawn from the supply in the external resistance R = I

I = $\frac{E}{R+r}$

on putting the values ,

 = $\frac{12.0V}{(8.5+0.09)ohm}$

= 1.4 A

let the terminal Voltage be V.

The terminal Voltage across the supply is the same as that across R.

V = IR                                  [by ohm's law]

  = 1.4 A * 8.5  Ω

 = 11.9 V

Answer 2

$\huge\textbf{Answer:-}$

2.0 V is the emf of a single cell

Total number of emf

6 cells

$= (6 ° 2.0 )= (12.0 V)$

0.015 is the resistance of a single cell

Total resistance

$(r = 6° ) \\ (0.015 = 0.09 )$

$Therefore,(R = 8.5 )$

$(R = I)$

$1 = \frac{E}{R + r}$

Adding the values :

$= \frac{12.0v}{8.5 + 0.9} \\ = 1.4$

(V=IR)

It's found by using the {ohm's Law}

$1.4a = 8.5 \\ 11.9v$

$\huge\textbf{Hop-it-Helps!!!}$