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In the circuit shown in figure, epsilon(...

In the circuit shown in figure, `epsilon_(1) = ` 3 V, ` epsilon_(2) = 2 V , epsilon_(3) = IV ` and ` R = r_(1) = r_(2) = r_(3) = 1 Omega` .
The current through each branch is...... and potential difference between the points A and B is .......

Text Solution

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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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