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In the circuit shown in figure, epsilon_(1) =3 V,epsilon_(2) = 2 V , epsilon_(3) = IVandR = r_(1) = r_(2) = r_(3)= 1 Omega . The current through each branch is...... and potential difference between the points A and B is ....... |
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Answer» Solution :Let `I_(1) ,I_(2) and I_(3)` be the currents through the resistors `r_(1), r_(2) and r_(3)`respectively as indicated in figure. USING Kirchoff.s second RULE to loops abaca and abcdefa we have, `I_(1) r_(1) - epsilon_(1) + epsilon_(2) + I_(2) r_(2) = 0 ""` ..... (1) `I_(1) r_(1) - epsilon_(1) + epsilon_(3)+ I_(3)r_(3) = 0""` ..... (2) From equation (1) and (2), `epsilon_(1) - I_(1) r_(1) = epsilon_(2) + I_(2) r_(2) = epsilon_(3) + I_(3) r_(3) ""` .... (3) Applying Kirchoff.s first rule to junction a, we have `I_(1)= I_(1) + I_(3) "" ` .... (4) Using equation (4) in (3) , we get , ` epsilon_(1) - (I_(2) + I_(3) ) r_(1) = epsilon_(3) + I_(3) r_(3)` or, `2I_(3) + I_(2) = 2"" `.... (5) Also `epsilon_(2) +I_(2) r_(2) = epsilon_(3) + I_(3) r_(3)` or , `I_(3) - I_(2) = I "" ` .... (6) From equation (4) , (5) and (6) , `I_(1)= IA, I_(2) = 0 A and I_(3) = 1 A ` Potential difference between A and B = Potential difference between a and d = `epsilon_(1) - I_(1) r_(1) = 3 -1 XX 1 = 2 ` V |
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