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The emf of the batteries E, F, G and H are 2 V, 1 V, 3 V and 1 V respectively. Their internal resistance are respectively 2 Omega, 1 Omega,3 Omega and 1 Omegarespectively. Calculate potential difference between B and D. |
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Answer» Solution :Circuit with internal resistances of batteries is as under. Currents on different PATHS are shown in figure. Near point D according to Kirchoff.s first LAW, I = `I_(1) + I_(2) ""` .... (1) For loop DBAD using Kirchoff.s second law, `2I_(1) - 2 (I - I_(1))- (I - I_(1))= -2 + 1 ` `THEREFORE 2I_(1) - 2I+ 2I_(1) - I + I_(1) = - 1` `therefore 5I_(1) -3I = -1""` ... (2)  For loop D-C-B-D using Kirchoff.s second law, `therefore- 3I - I - 2I_(1) = - 3 + 1 ` `therefore - 4I- 4I_(1) = - 2 ` ` therefore 2 I + I_(1) = 1"" ` .... (3) Multiplying equ. (2) by 2 and equ. (3) by 3 and eliminating l , ` therefore 10I_(1) - 6I = -2 ` `6I + 3I_(1) = 3 ` `therefore 13 I_(1) = 1 rArr I_(1) = (1)/(13) `A p.d. between points B and D=` V_(BD) = 2I_(1)` ` therefore V_(BD) = (2)/(13) `V |
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