Question

$1.$ Which student measurement is wrong in table B?
$2.$ What is the mathematical relation between voltage and current?
$3.$ In the following measurement of student B. Which measurement is wrong?
$4.$ The value of resistance from the measurement of student A is?

Student	S.No	Voltmeter Reading(mv)	Ammeter Reading(mA)
Student A	$1$ $2$ $3$	$2$ $4$ $6$	$1$ $2$ $3$
Student B	$1$ $2$ $3$	$4$ $6$ $8$	$4$ $3$ $4$

(a) $I = 4,V = 4$
(b) $I = 3,V = 6$
(c) $I = 4,V = 8$
(d) None of these

Answer 1

In order to solve this question we need to understand the development of resistance in circuits. So when a wire is subjected to potential difference then an electric field setup inside wire which bounds the free electrons to move in opposite side as of applied electric field so electric current flows in direction opposite to flow of electrons or in direction to the decreasing potential. When a current flows inside wire then wire opposes this behavior and shows or develops opposition known as resistance of wire.

Complete answer:
Part( $1$ ): The student which measures wrong readings is student B because as we can see that the ratio of voltage to current is not same in all three readings done by student B as $\dfrac{4}{4} = 1$ in first reading whereas second reading has a ratio of $\dfrac{6}{3} = 2$ Hence, these readings are wrong according to the ohm’s law which says $\dfrac{V}{I} = R$ the ratio of voltage and current must be constant which is known as resistance.
Part( $2$ ): Ohm’s law for metals is stated the potential difference across the wire is directly proportional to the current produced in wire, that is $V \propto I$
And the proportionality constant is known as, $R$ Resistance
So Ohm’s law is mathematically stated as, $V = RI$
Or, $R = \dfrac{V}{I}$
Part( $3$ ): The only reading of student B will be wrong if the ratio of voltage and current is not as same of other two readings and in part(a) the ratio of voltage to current is $\dfrac{4}{4} = 1$ in part(b) the ratio of voltage to current is $\dfrac{6}{3} = 2$ and in part(c) the ratio is $\dfrac{8}{4} = 2$ hence the part(a) has wrong reading because the ratio is different from other two correct readings.
Part( $4$ ): Resistance in each of three reading by student A is calculated by taking ratio of voltage and current readings and since the ratio of each reading by student A is same which is $\dfrac{2}{1} = \dfrac{4}{2} = \dfrac{6}{3} = 2$ and from ohm’s law the ratio of voltage and current is the resistance hence, Resistance measured by student A is $2\Omega$ .

Note:
It should be remembered that resistance is not constant at every temperature, rather it increases with temperature for metals but for semiconductors it decreases with resistance because semiconductors have a negative coefficient of resistance. So a semiconductor behaves as a conductor at high temperature due to low value of electrical resistance.