High School

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A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be :-
A.3 v
B.1.5 v
C.1 v
D.2 v

Answer :

The potential difference between two points separated by 50 cm on a wire with a uniform resistance of 100 ohms connected to a 6 volt battery is 1 volt (C.1 v). We find this by calculating the resistance per length of wire and the current, and then applying Ohm's law to the 50 cm section of wire. Thus the correct answer is C.1 v.

To determine the potential difference between two points on the wire, we can use Ohm's law (V=IR), where V is the voltage, I is the current, and R is the resistance. The voltage drop across a certain length of wire can be calculated if we know the total resistance and current in the circuit. Since the wire has a uniform thickness and resistance, the resistance per unit length will also be uniform. The total resistance of the wire is 100 ohms, and it is 3 meters long, so the resistance per meter is approximately 33.33 ohms.

Using the battery voltage of 6 volts and the total resistance, the current flowing through the wire can be calculated as

[tex]I = V/R,[/tex]

which is

[tex]6V / 100 ohms = 0.06 A.[/tex]

Then, the potential difference across 0.5 meters (50 cm) can be found by multiplying this current by the resistance for that length of wire (50 cm of wire has approximately 16.67 ohms of resistance, since 33.33 ohms is for 1 meter).

Therefore, the potential difference for 50 cm of wire is

[tex]V = IR = (0.06 A)(16.67 ohms) = 1 V.[/tex]

Hence, the correct answer is C.1 v. Thus the correct answer is C.1 v.

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