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A cell of e.m.f. E and internal resistance r is connected in series with an external resistance nr.Then,the ratio of the terminal potential difference to e.m.f. is a.1/n b.1/(n+1) c.n/(n+1) d.(n+1)/n

A cell of e.m.f. E and internal resistance r is connected in series with an external resistance nr.Then,the ratio of the terminal potential difference to e.m.f. is


a.1/n


b.1/(n+1)


c.n/(n+1)


d.(n+1)/n

Grade:12th Pass

2 Answers

Ashwin Muralidharan IIT Madras
290 Points
12 years ago

Hi Menka,

 

E = I(nr+r) = Ir(n+1).

Terminal PD = Inr.

 

Hence the ratio = n/(n+1). Option (c).

 

Hope this helps.

 

Regards,

Ashwin (IIT Madras).

Sagnik Bairagya
30 Points
4 years ago
We know that in series combination, 
I = E net / (r net + R)
or, I * (r net + R) = E net
So, placing values as per question we get :-
E = I * (r  + nr)  …..... First part.
 
Now,
We also know the formula for terminal potential difference.
i.e.  V = E - I r
Now substituting the value of E we get
V= Ir - Inr -Ir
or, V = Inr
 
Therefore,
ratio:-
p.d / E.M.F = Inr / { Ir (1+n) } 
= n / (n+1)  which is the correct answer...
 
Hope that will help.
 

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