Answer to Question #184778 in Electric Circuits for Bernadette Akwe

Question

A battery made up of 2cell joined in series supplies current to an external resistance of 5ohms.if the emf and internal resistance of each cell is o.6v and 6ohms respectively calculate the current flowing in the external resistor (b)calculate the terminal p.d. (d) calculate the lost voltage and construct a circuit

Answer 1

(a)

using kvl

0.6-5I-6I+0.6-6I+0.6=0

$\implies$ $\boxed{I = \frac{1.2}{17} A}$

this is the current that will flow through the external resistor

(b)

Terminal Voltage = emf - Voltage drop

emf = 1.2

terminal voltage of equvalent of both the batteries is as equvalent emf is 1.2V and equvalent internal resistence is 12 ohm (using V = v₁+v₂ , r = r₁+r₂ for series battery )

= $1.2-\frac{1.2}{17}\times12$

=0.36V

(c)/(d)

voltage loss is Ir (here r is equivalent internal resistance of batteries as batteries are in series and now combined to one with equivalent resistance of 12 ohm )

= $\frac {1.2}{17}\times12$

= 8.4V