1998/II/33 PHY CE

The crucial point why option E is wrong is because the supply voltage is unchanged doesn't mean that the voltage across the long connecting wire is also unchanged.

Since the wire and the appliance consitute a series circuit, the supplied volatge is shared both by the wire and appliance according to the equation:

V = E.[R/(R+R')]
where V is the voltage across the wire
E is the supplied voltage
R is the resistance of the wire, and R' is the resistance of the appliance.

When R is reduced to R/2, the new voltage across the wire V' becomes
V' = E[R/2(R/2+R')], which is clearly not equal to V.

Ohm's Law shows that halving the resistance of a component leads to doubling of the current only under the condition that the voltage across that component is constant. Now, becuase the voltage across the wire is not constant, the current will not be doubled.

The correctness of option C is apparent, as shown by Ohm's Law.

Let the resistance of the electrical appliance be r.
By V=IR
V=(1/2(I))(r+R)
2V=I(r+R)
So,if the total resistance of the connecting wires remains unchanged, then doubling the voltage will double the current.
If the voltage of powe supply remains unchanged and reducest otal resistance of conneting wires to R/2, the current will not change to I.