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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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Answer» 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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