Potential difference is doubled and resistivity is halved what will happen to current
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According to Ohm's law
V= I × R
where , V is the potential difference
I is the current
R is the resistance
so I= V÷R
now resistance (R) proportional to resistivity (roh)
if resistivity is halved then the resistance R will also halved.
new resistance R'=R/2
new potential difference = V'=2V
so new current
I' = V'÷R'
=2V÷(R/2)
=4V÷R
= 4 × I
so the current become 4 times of its initial value.
V= I × R
where , V is the potential difference
I is the current
R is the resistance
so I= V÷R
now resistance (R) proportional to resistivity (roh)
if resistivity is halved then the resistance R will also halved.
new resistance R'=R/2
new potential difference = V'=2V
so new current
I' = V'÷R'
=2V÷(R/2)
=4V÷R
= 4 × I
so the current become 4 times of its initial value.
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