Question 521: Current Electricity

Question

A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance $100 \Omega$. The difference of potential between two points 50 cm apart on the wire is

Accepted Answer

Resistance per unit length
:
$    R = \frac{ho l}{A}    $
Given that the total resistance of the 3 m wire is 100 ÃÂ©, we can express the resistance per unit length as:
$    \frac{100}{3} = \frac{ho}{A}    $
Therefore, the resistance for 1 meter of the wire is $\frac{100}{3} \, \Omega/	ext{m}$.
Resistance of a 50 cm segment
:
The length of the segment in meters is 0.5 m. The resistance of this segment is:
$    R' = \frac{ho}{A} 	imes l = \frac{100}{3} 	imes 0.5 = \frac{50}{3} \, \Omega    $
Current through the wire
:
The current $I$ flowing through the entire wire when connected to a 6 V battery can be calculated using Ohm's law:
$    I = \frac{V}{R} = \frac{6}{100} \, 	ext{A}    $
Potential difference across the 50 cm segment
:
Using the formula $V = IR'$, we find the potential difference between the two points 50 cm apart:
$    V = I 	imes R' = \frac{6}{100} 	imes \frac{50}{3} = 1 \, 	ext{V}    $
Thus, the potential difference between the two points 50 cm apart on the wire is $1 \, 	ext{V}$.

Answer

1 V

Answer

1.5 V

Answer

2 V

Answer

3 V

Current Electricity

A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance 100 \Omega. The difference of potential between two points 50 cm apart on the wire is

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