Question

In a series electrical circuit comprising a resistor made up of metallic wire, ammeter reads 100 ma . If the light of the wire is doubled how will the current in the circuit change

Answer 1

Explanation:

i guess it will be clear after seeing this

Answer 2

Answer:  yes, there is a change in current ,it will be half

                current = 50 ma

Explanation:

• The resistance doubles with a twofold increase in wire length. because the relationship between resistance and wire length is linear.
• Consequently, under Ohm's Law for a specific potential difference.
• Half of the current is reduced.
•                              $I$$\alpha$$1/R$
• Before being doubled in length.
•                           $I=V/R= 100mA$
• Resistance doubles when length does as well.
•                               $I=V/2RI=I/2I=100mA/2\\I=50mA$
• Ohm's Law:
•       Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, presuming that all physical characteristics and temperatures remain constant.
•                                  $V=IR$
• The proportionality constant, denoted by the initials R in the equation, has ohms as its unit of measurement.
• To calculate the resistance and current, respectively, the same formula can be recast as follows:
•                                       $I=V/R\\R=V/I$.
• This is demonstrated by the filament of a light bulb, where the temperature rise as the current rises. Ohm's law is inapplicable not this circumstance. A lightbulb's filament disobeys Ohm's Law.
• Ohm’s Law Statement: When the temperature and all other physical factors are held constant, Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it.

• Ohm’s Law Equation: V is the voltage applied across the conductor, I is the current flowing through the conductor, and R is the resistance the conductor offers to the passage of current. These three variables together form the equation V = IR.

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