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In the network shown in given figure , ...

In the network shown in given figure , find the potential differene across BD.

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Resistance of branch ABC ` = (10+5) Omega = 15 Omega`
Resistance of branch ADC ` = (10 + 5) Omega = 15 Omega`
Both branches are parallel across the cell 3 V.
Current through both branches are
`l_(1) = (3V)/(15 Omega) = (1)/(5) A and l_(2) = (3V)/(15 Omega) = (1)/(5)A`

Potential difference across BD
`V_(B) - V_(D) = (V_(B) - V_(A)) + (V_(A) - V_(D))`
` = - 5l_(1) + 10l_(2)`
` = - 5 xx (1)/(5) xx 10 + (1)/(5)`
= 1 V
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