Ohm's Law
This law,
which is applicable to electrical conduction through reliable conductors, can
be formulated as follows.
If the
conductor's temperature is constant, the ratio of the potential difference (V)
between any two places on it to the current (I) flowing between them will
remain constant.
What I mean is, V/I = constant, or V/I = R
where R is
the resistance of the conductor between the two points considered.
Put in
another way, It simply means that provided R is kept constant, current is
directly proportional to the potential difference across the ends of a
conductor. However, this linear relationship between V and I does not apply to
all non-metallic conductors. For example, for silicon carbide, the relationship
is given by V = KIm where K
and m are constants and m is less
A coil of
copper has a resistance of 20 ohm at 10oC and is connected to a 220 V supply. By how
much must the voltage be Increased in order to maintain current constant if the
temperature of the coil rises to 40oC? Take the tepmperature
coefficient of resistance of copper as 0.00428 from 0oC
Solution:
first, we
have to use this formula to calculate the resistance of the conductor after
changing the temperature:
R40 /R10
= (1 + 40 x 0.00428) / (1 + 10 x 0.00428)
R40 =
20 x 1.1712/1.0428 = 22.5 ohm
After that, calculate the current at the initial temperature, which is at a temperature of 10oC
V/I = R
I = V/R
I = 220/20 =
11 Ampere
Since, the wire resistance has become 22.5 Ohm at 40oC, the new voltage required for keeping the current constant at its previous value is
V/I = R
V = I x R =
11 x 22.5 = 247.5
Increase
in voltage required = 220 – 247.5 = 27.5
Volt
A coil of
copper has a resistance of 20 ohm at 50oC and is connected to a 240 V supply. By how
much must the voltage be Decreased in order to maintain current constant if the
coil temperature drops to 30oC ? Take the tepmperature coefficient
of resistance of copper as 0.00428 from 0oC
first, we
have to use this formula to calculate the resistance of the conductor after
changing the temperature:
R30 /R50
= (1 + 30 x 0.00428) / (1 + 50 x 0.00428)
R30 =
20 x 1.1284/1.214 = 18.6 ohm
After that, calculate the current at the initial temperature, which is at a temperature of 50oC
V/I = R
I = V/R
I = 240/20 =
12 Ampere
Since, the wire resistance has become 18.6 Ohm at 30oC, the new voltage required for keeping the current constant at its previous value is
V/I = R
V = I x R =
12 x 18.6 = 223.2
Decrease
in voltage required = 240 – 223.2 = 16.8
Volt
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