How do you think the total resistance of a parallel circuit is calculated?
1 answer:
Let's consider two resistors only for simplicity. When the resistors are connected in parallel, the potential difference on each resistor is the same, and it is equal to the voltage of the battery, V:
(1)
Instead, the current splits in the two resistors:
(2)
where I is the current in the circuit, I1 and I2 are the currents in the two resistors.
By using Ohm's law, we can rewrite (2) as
(3)
where
is the equivalent resistance of the circuit. But we also said that
So we can rewrite (3) as
which becomes
So, the equivalent resistance of the parallel of n resistors is given by
Instead, the current splits in the two resistors:
where I is the current in the circuit, I1 and I2 are the currents in the two resistors.
By using Ohm's law, we can rewrite (2) as
where
So we can rewrite (3) as
which becomes
So, the equivalent resistance of the parallel of n resistors is given by
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The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
where
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.
The relationship between the resistance and the resistivity of a wire is
where
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.