Answer to Question #101925 in Electric Circuits for jude

Question

A cell with emf 15V and internal resistance 5.6 Ohms is connected in series to an ammeter and a 0-3.9MegaOhm variable resistor. A voltmeter is connected across the terminals of the cell..
a) Sketch the circuit diagram
b) Describe how this circuit can be used to obtain a value for the emf of the cell
c) Sketch and label the graph
d) Calculate the terminal voltage when a current of 87.9 microAmps is measured by the ammeter
e) Calculate the load resistance when a current of 55.8 microAmps is measured by the ammeter
f) Describe several limitations of this investigation

Answer 1

As per the given question,

The emf of the cell =15V

Internal resistance of the cell =5.6 ohm

External resistance R=$0-3.9\times 10^6 Ohm$

a)

b) We know that V= E- i R

By putting the value of i, the potential difference can be calculated and the the current losses in the internal resistance,

equivalent resistance $R_{eq}=R_{load}+5.6 ohm$

Let the current in the circuit is i,

Now, applying the kirchoff rule,

$\Rightarrow V=iR_{eq}$

$\Rightarrow 15=iR+i\times 5.6$

$\Rightarrow i=\dfrac{15}{R+5.6}$ or $R=\dfrac{15}{i}-5.6$





d) As per the question current 87.9 microAmps

So, terminal voltage $V= iR_{load}=87.9\times10^{-6}\times3.9\times10^{6}=342.81V$

e)

$i= 55.8 \mu A$

$\Rightarrow i=\dfrac{15}{R+5.6}$

$\Rightarrow R=\dfrac{15-5.6\times55.8\times10^{-6}}{55.8\times10^{-6}}$

$\Rightarrow R=\dfrac{14.9996}{55.8\times10^{-6}}$ $=0.26\times 10^{6} ohm$

f) It can not be applied for the unilateral circuits.

It is not applicable for the non linear elements.