Question

The resistance of a conductor is reduced to half its initial value. In doing so the heating effects in the conductor will become :

a) Half

b) One-fourth

c) Four times

d) Double

Answer 1

@Fifciol

Answer 2

@Festinger

Answer 3

The heating effect of a conductor is given by:
\[P=VI\]
Assume the conductor is ohmic and thus obeys ohm's law:
\[V=IR\]

Putting this in the first question you can see that
\[P=I^{2}R\]

Before making any changes,
\[V=I_{1}R_{1}\]
and so power is
\[P_{1}=\frac{V^{2}}{R_{1}}\]

But after we reduce the resistance, current will be another value, following ohm's law:
\[V=I_{2}R_{2}\]
But
\[R_{2}=\frac{1}{2}R_{1}\]
and so
\[V=\frac{I_{2}R_{1}}{2}\]
And thus
\[I_{2}=\frac{2V}{R_{1}}\]
and power is then:
\[P_{2}=I^{2}R_{2}=\frac{4V^{2}R_{1}}{2R_{1}^{2}}=\frac{2V^{2}}{R_{1}}\]

Comparing with the first result the heating effect doubles.

Answer 4

the heat effect increases because
voltage is directly proportional to current as there it there is some heat but when the
resistance of a conductor is reduced to half its initial value then the heat become
d) doubled
for example: the high voltage fire the materials.

Answer 5

power dissipiated as heat=v^2/r.. if r is halved.. heat produced doubles

Answer 6

\[P _{old} = I _{old} ^{2}R _{old}\]
\[I _{new}=\frac{ V }{ R _{old}/2 }=\frac{ 2V }{ R _{old} }=2I _{old}\]
\[P _{new}=I _{new}^{2}R _{new}=2^{^2} I _{old }^{2}R _{old}/2=\frac{ 4 }{ 2 } I _{old }^{2}R _{old}=2P _{old}\]
It doubles.

Answer 7

thank you so much guys....for helping me
@Festinger @Imtiaz7 @mr.singh @ybarrap