A `16 Omega` resistance wire is bent to form a square. A source of emf 9 V is connected across one of its sides as shown. Calculate the current drawn from the source. Find the potential difference between the ends C and D. If now the wire is stretched uniformly to double the length and once again the same cell is conected in the same way, across one side of the square formed, what will now be the potential difference across one of its diagonals?

Explanation

Their equivalent resistance is given by
`(1)/(R_("eff")) = (1)/(R.) + (1)/(R_(AB)) = (1)/(12) + (1)/(4)`
`R_("eff") = 3Omega`
The current supplied by the source to the network: `I = (V)/(R_("eff")) = (9)/(3) = 3A`
Potential drop across AB= 9V
`rArr I. xx R_(AB) = 9 rArr I" xx 4= 9`
`rArr I. = (9)/(4) A`
Potential drop across `CD= (I-I.) xx 4 Omega`
`=(3- (9)/(4)) xx 4= 3V`
The wire is stretched to double of its length keeping volume constant, we have
`R= rho (l)/(A)`
`rArr R= rho (l^(2))/(Al)`
Since volume = A.l is constant, we have
`R prop l^(2)`
Since `l_("new") = 2l`
`:. R_("new") = 4R= 4 xx 4 = 16 Omega`
The new effective resistance of their network `R_("eff") = (48 xx 16)/(64) = 12 Omega`
Current in this network
`I_(1) = (V)/(R_("eff")) = (9)/(12) = (3)/(4)A`
Current `I_(2), 16 xx I_(2) = 9`
`rArr I_(2) = (9)/(16)A`
Potential across the diagonal A to C `=(I_(1) - I_(2)) xx(16 +16)`
`=((3)/(4) -(9)/(16)) xx 16 = 3V`