Dear student Emf of driving cell=Eresistance of potentiometer wire is R and internal resistance of cell be r so current through potentiometer wire at null point that is when galvanometer reading is zero is i=ER+rLet balancing length be l so potential drop across it at null point is V=i×ρlA=ER+r×ρlA (ρ is resisvity of wire and A is its area of cross section)At null point potential drop across the balancing length of potentiometer wire =Emf of the cell cennected in parallel across it E1=ER+r×ρlAl=E1×(r+R)×AE×ρinternal resistance of driving cell r increases than balancing length also increasesRegards

ii part

Dear student

E m f o f d r i v i n g c e l l = E r e s i s \tan c e o f p o t e n t i o m e t e r w i r e i s R a n d i n t e r n a l r e s i s \tan c e o f c e l l b e r s o c u r r e n t t h r o u g h p o t e n t i o m e t e r w i r e a t n u l l p o i n t t h a t i s w h e n g a l v a n o m e t e r r e a d i n g i s z e r o i s i = \frac{E}{R + r} L e t b a l a n c i n g l e n g t h b e l s o p o t e n t i a l d r o p a c r o s s i t a t n u l l p o i n t i s V = i \times \frac{ρ l}{A} = \frac{E}{R + r} \times \frac{ρ l}{A} (ρ i s r e s i s v i t y o f w i r e a n d A i s i t s a r e a o f c r o s s s e c t i o n) A t n u l l p o i n t p o t e n t i a l d r o p a c r o s s t h e b a l a n c i n g l e n g t h o f p o t e n t i o m e t e r w i r e = E m f o f t h e c e l l c e n n e c t e d i n p a r a l l e l a c r o s s i t . E_{1} = \frac{E}{R + r} \times \frac{ρ l}{A} l = \frac{E_{1} \times (r + R) \times A}{E \times ρ} i n t e r n a l r e s i s \tan c e o f d r i v i n g c e l l r i n c r e a s e s t h a n b a l a n c i n g l e n g t h a l s o i n c r e a s e s R e g a r d s

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