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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. isa.1/nb.1/(n+1)c.n/(n+1)d.(n+1)/n

9 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,

2 years ago
We know that in series combination,
I = net / (r net + R)
or, I * (r net + R) = net
So, placing values as per question we get :-
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.