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+1 vote
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in Current Electricity by (20 points)
edited by

Two cells of emf 3 V and 4 V and internal resistance 1 omega and 2 omega respectively are connected in parallel so as to send the current in the same direction through an external resistance of 5Q. Find the potential difference across 5Q resistor.

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1 Answer

+1 vote
by (32.8k points)

Given : r1 = 1Ω

r2 = 2 Ω

E1 = 3v

E2 = 4v

Total equivalent internal resistance 

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{r_1}\) + \(\cfrac{1}{r_2}\)

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{1}\) + \(\cfrac{1}{2}\)

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{3}{2}\)

req\(\cfrac{2}{3}\)Ω

Equivalent emf of cell in parallel connection

Eeq = \(\left[\cfrac{E_1}{r_1}+\cfrac{E_2}{r_2}\right]r_{eq}\) 

Eeq = \(\left[\cfrac{3}{1}+\cfrac{4}{2}\right]\times \cfrac23\)  

 Eeq = \(\left[\cfrac{6+4}2\right]\times \cfrac23\) 

 Eeq = 5 x \(\cfrac{2}{3}\)

 Eeq = 3.3 v

potential difference across 5 Ω resistor

E = \(\left[\cfrac{R}{R+r_{eq}}\right]\)  Eeq

E = \(\left[\cfrac{5}{5+\cfrac23}\right]\) x 3.3

\(\cfrac{15}{17}\) x 3.3

= 2.9 v

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