Ohm's Law and Resistance Questions in English

(N/A) The resistance of a conductor depends upon the following factors:
$(a)$ Length of the conductor: Resistance is directly proportional to the length $(R \propto l)$.
$(b)$ Cross-sectional area of the conductor: Resistance is inversely proportional to the cross-sectional area $(R \propto 1/A)$.
$(c)$ Material of the conductor: Resistance depends on the nature of the material, represented by its resistivity $(\rho)$.
$(d)$ Temperature of the conductor: For most conductors, resistance increases with an increase in temperature.

(A) The resistance of a wire is given by the formula $R = \rho \frac{l}{A}$.
Where:
$\rho$ = Resistivity of the material of the wire.
$l$ = Length of the wire.
$A$ = Area of cross-section of the wire.
From the formula,it is clear that resistance is inversely proportional to the area of cross-section $(R \propto \frac{1}{A})$.
$A$ thicker wire has a larger area of cross-section,which results in lower resistance. Conversely,a thinner wire has a smaller area of cross-section,resulting in higher resistance.
Since current flows more easily through a path of lower resistance,current will flow more easily through a thick wire compared to a thin wire when connected to the same source.

(C) According to Ohm's law, the relationship between potential difference $(V)$, current $(I)$, and resistance $(R)$ is given by:
$V = IR$ or $I = \frac{V}{R}$
Given that the resistance $(R)$ remains constant, the current is directly proportional to the potential difference $(I \propto V)$.
If the potential difference is reduced to half of its former value, let the new potential difference be $V' = \frac{V}{2}$.
Substituting this into the formula for the new current $(I')$:
$I' = \frac{V'}{R} = \frac{\frac{V}{2}}{R} = \frac{1}{2} \left( \frac{V}{R} \right) = \frac{I}{2}$
Therefore, the current flowing through the electrical component is reduced to half of its original value.

(A) The electrical resistivity of iron is $10.0 \times 10^{-8} \,\Omega \,m$.
The electrical resistivity of mercury is $94.0 \times 10^{-8} \,\Omega \,m$.
Since the resistivity of mercury is higher than that of iron, mercury offers more resistance to the flow of electric current.
Therefore, iron is a better conductor than mercury.

(A) The electrical conductivity of a material is inversely proportional to its electrical resistivity.
Materials with lower resistivity are better conductors of electricity.
Comparing the values in the table,silver has the lowest resistivity of $1.6 \times 10^{-8} \, \Omega \, m$.
Therefore,silver is the best conductor among the given materials.

(A) The resistance $(R)$ of a copper wire of length $l$ and cross-sectional area $A$ is given by the formula:
$R = \rho \frac{l}{A}$
Given:
Resistivity of copper,$\rho = 1.6 \times 10^{-8}\, \Omega \, m$
Diameter,$d = 0.5\, mm = 5 \times 10^{-4}\, m$
Resistance,$R = 10\, \Omega$
Area of cross-section,$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5 \times 10^{-4}}{2}\right)^2 = \pi \times 6.25 \times 10^{-8}\, m^2$
Calculating length $(l)$:
$l = \frac{R \times A}{\rho} = \frac{10 \times 3.14 \times 6.25 \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{196.25}{1.6} \approx 122.7\, m$
If the diameter is doubled,the new diameter $d' = 2d = 1.0\, mm = 10^{-3}\, m$.
Since $R \propto \frac{1}{d^2}$,if the diameter is doubled,the area $A$ becomes $4$ times larger.
Therefore,the new resistance $R' = \frac{R}{4} = \frac{10}{4} = 2.5\, \Omega$.

(N/A) The plot between voltage and current is called $IV$ characteristic. The voltage is plotted on the $x-$axis and current is plotted on the $y-$axis.
The $IV$ characteristic of the given resistor is plotted in the figure.
The slope of the line gives the value of $\frac{1}{R}$ as,
Slope $= \frac{1}{R} = \frac{\Delta I}{\Delta V} = \frac{BC}{AC} = \frac{3.0 - 1.0}{10.2 - 3.4} = \frac{2.0}{6.8}$
Therefore,the resistance $R$ is,
$R = \frac{6.8}{2.0} = 3.4\, \Omega$
Thus,the resistance of the resistor is $3.4\, \Omega$.

(B) Resistance $(R)$ of a resistor is given by Ohm's law as:
$V = IR$
$R = \frac{V}{I}$
Given:
Potential difference,$V = 12\, V$
Current in the circuit,$I = 2.5\, mA = 2.5 \times 10^{-3}\, A$
Substituting the values in the formula:
$R = \frac{12}{2.5 \times 10^{-3}} = \frac{12}{0.0025} = 4800\, \Omega$
$R = 4.8 \times 10^3\, \Omega = 4.8\, k\Omega$
Therefore,the resistance of the resistor is $4.8\, k\Omega$.

(C) The electrical resistivity $(\rho)$ of a material is an intrinsic property that depends only on the nature of the material and the temperature of the conductor.
It does not depend on the physical dimensions of the wire, such as its length, thickness (cross-sectional area), or shape.
While resistance $(R)$ depends on length and area $(R = \rho \frac{l}{A})$, resistivity $(\rho)$ remains constant for a given material at a specific temperature.

(B) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity of the material,$l$ is the length,and $A$ is the area of cross-section.
For the first conductor: $R = \rho \frac{l}{A}$.
For the second conductor,the length is $2l$,the resistance is $R$,and the material is the same (so $\rho$ remains constant). Let the new area be $A'$.
Then,$R = \rho \frac{2l}{A'}$.
Equating the two expressions for $R$: $\rho \frac{l}{A} = \rho \frac{2l}{A'}$.
Canceling $\rho$ and $l$ from both sides: $\frac{1}{A} = \frac{2}{A'}$.
Solving for $A'$,we get $A' = 2A$.

(C) According to Ohm's law, $V = IR$, which implies $I = V/R$ or $R = V/I$.
In the given $I-V$ graph, the slope of the line represents $I/V$, which is equal to $1/R$.
Therefore, the slope of the graph is inversely proportional to the resistance $(Slope \propto 1/R)$.
From the figure, the slope of the line for $R_1$ is the greatest, followed by $R_2$, and then $R_3$ (i.e., $Slope_1 > Slope_2 > Slope_3$).
Since the slope is inversely proportional to resistance, a higher slope indicates a lower resistance.
Thus, $R_1 < R_2 < R_3$.

(A) Resistivity $(\rho)$ is an intrinsic property of a material.
It depends only on the nature of the material and the temperature.
It does not depend on the physical dimensions (length, area of cross-section) or the shape of the resistor.
Therefore, if the shape of the resistor is changed, the resistance $(R)$ will change, but the resistivity $(\rho)$ remains constant.

(N/A) The errors in the given circuit diagram are as follows:
$1$. The ammeter $(A)$ is connected in parallel with the resistor $(R)$,whereas it should be connected in series with the resistor.
$2$. The voltmeter $(V)$ is connected in series with the circuit,whereas it should be connected in parallel across the resistor $(R)$.
$3$. The polarity of the ammeter and voltmeter is incorrect in the diagram.
Correct circuit configuration:
$1$. Connect the ammeter $(A)$ in series with the resistor $(R)$.
$2$. Connect the voltmeter $(V)$ in parallel across the resistor $(R)$.
$3$. Ensure the positive terminal of the ammeter and voltmeter is connected towards the positive terminal of the battery,and the negative terminal is connected towards the negative terminal of the battery.

(A) The resistance of an ammeter should be as low as possible,ideally $0 \ \Omega$.
An ammeter is connected in series with the circuit to measure the current flowing through it.
If the resistance of the ammeter is high,it will significantly reduce the total current in the circuit due to the increased total resistance $(R_{total} = R_{circuit} + R_{ammeter})$.
Therefore,to ensure the ammeter measures the true current without altering the circuit's behavior,its resistance must be negligible.

(B) Electrical resistivity is the resistance offered by a unit length and unit cross-sectional area of a material. It is a characteristic property of the material.
According to the formula $R = \rho \frac{l}{A}$, where $R$ is resistance, $\rho$ is resistivity, $l$ is length, and $A$ is the cross-sectional area.
When the length of the wire is doubled $(l' = 2l)$, the resistance becomes $R' = \rho \frac{2l}{A} = 2R$.
Since the voltage $V$ remains constant in the circuit, according to Ohm's Law $(V = IR)$, the current $I$ is inversely proportional to resistance $(I = \frac{V}{R})$.
Therefore, when the resistance is doubled, the current becomes half of its original value, i.e., $I' = \frac{V}{2R} = \frac{I}{2} = \frac{5}{2} = 2.5\, A$.

(N/A) Ohm's law states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ across its ends,provided the temperature and other physical conditions remain constant. Mathematically,$V \propto I$ or $V = IR$,where $R$ is the resistance of the conductor.
Experimental Verification:
$1$. Set up a circuit consisting of a resistor,an ammeter in series,a voltmeter in parallel across the resistor,a battery,and a rheostat.
$2$. Adjust the rheostat to change the current in the circuit and record the corresponding readings of the ammeter $(I)$ and voltmeter $(V)$.
$3$. Plot a graph of $V$ (y-axis) versus $I$ (x-axis). $A$ straight line passing through the origin confirms Ohm's law.
Limitations:
Ohm's law does not hold under all conditions. It is not applicable to non-ohmic devices like diodes,transistors,and vacuum tubes,where the $V-I$ relationship is non-linear. Additionally,for ohmic conductors,if the temperature changes significantly due to heating,the resistance $R$ changes,and the law may not hold strictly.

(N/A) Electrical resistivity is defined as the resistance offered by a material of unit length and unit cross-sectional area. Mathematically,$R = \rho (l/A)$,where $\rho$ is the resistivity. Its $SI$ unit is ohm-metre $(\Omega \, m)$. To study the factors affecting resistance: $1$. Take a battery,an ammeter,a plug key,and nichrome wires of different lengths but same thickness. $2$. Connect the wire in the circuit and measure the current using the ammeter. $3$. Observe that resistance increases with the length of the wire $(R \propto l)$. $4$. Repeat the experiment using wires of the same length but different cross-sectional areas. $5$. Observe that resistance decreases as the area of cross-section increases $(R \propto 1/A)$.

(N/A) The resistance $(R)$ of a conductor depends on the following factors:
$1$. Length of the conductor $(l)$: Resistance is directly proportional to the length of the conductor $(R \propto l)$.
$2$. Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section of the conductor $(R \propto 1/A)$.
$3$. Nature of the material: Resistance depends on the resistivity $(\rho)$ of the material.
$4$. Temperature: Resistance of a conductor changes with a change in temperature.

(B) The resistivity of an alloy is generally much higher than that of its constituent pure metals.
This occurs because the crystal lattice of an alloy is distorted by the presence of different types of atoms,which increases the scattering of electrons,thereby increasing the electrical resistance.

(A) The $SI$ unit of resistivity is the ohm-metre, denoted as $\Omega \cdot m$.
Resistivity $(\rho)$ is defined by the formula $R = \rho \frac{l}{A}$, where $R$ is resistance, $l$ is length, and $A$ is the cross-sectional area.
Rearranging for resistivity gives $\rho = \frac{R \cdot A}{l}$.
Substituting the $SI$ units: $\Omega \cdot \frac{m^2}{m} = \Omega \cdot m$.

(B) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area of the wire.
From this relation,it is clear that the resistance $R$ is inversely proportional to the cross-sectional area $A$ $(R \propto \frac{1}{A})$.
Therefore,as the cross-sectional area increases,the resistance of the wire decreases.

(N/A) Ohm's law states that the electric current flowing through a metallic conductor is directly proportional to the potential difference $(V)$ applied across its ends,provided that its physical conditions,such as temperature,remain constant.
Mathematically,this is expressed as $V \propto I$ or $V = IR$,where $R$ is the constant of proportionality known as resistance.

(N/A) Resistance is the property of a conductor to oppose the flow of current (electrons) through it. It is defined as the ratio of the potential difference $(V)$ applied across the conductor to the current $(I)$ flowing through it, expressed by the formula $R = V/I$. The $SI$ unit of resistance is the ohm $(\Omega)$.

(C) The $SI$ unit of electrical resistance is the ohm,denoted by the Greek letter omega $(\Omega)$.
Resistance is defined as the opposition to the flow of electric current in a conductor.
According to Ohm's Law,$R = V / I$,where $R$ is resistance in ohms,$V$ is potential difference in volts,and $I$ is current in amperes.

(N/A) The resistance of a conductor is said to be $1$ ohm if a current of $1$ ampere flows through it when a potential difference of $1$ volt is applied across its ends. This is based on Ohm's Law, which states $R = V / I$. Therefore, $1 \Omega = 1 \ V / 1 \ A$.

(B) According to Ohm's Law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance of the conductor.
If the resistance $R$ remains constant,then $I = V/R$.
If the potential difference $V$ is doubled $(V' = 2V)$,the new current $I'$ becomes $I' = V'/R = (2V)/R = 2(V/R) = 2I$.
Therefore,the current flowing through the conductor is also doubled.

(D) The resistance $(R)$ of a conductor depends on the following factors:
$(i)$ Length of the conductor $(l)$: Resistance is directly proportional to the length of the conductor $(R \propto l)$.
$(ii)$ Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$(iii)$ Nature of the material: Resistance depends on the resistivity $(\rho)$ of the material.
$(iv)$ Temperature: Resistance of a conductor generally increases with an increase in temperature.

(C) The resistivity $(\rho)$ of a material is an intrinsic property that depends on the following factors:
$1$. Nature of the material: Different materials have different atomic structures, which affect the mobility of charge carriers.
$2$. Temperature: For most conductors, resistivity increases with an increase in temperature due to increased lattice vibrations, which hinder the flow of electrons.

(B) According to Ohm's Law,$V = IR$,which can be rearranged as $I = V/R$.
If the resistance $R$ is doubled $(R' = 2R)$ while the voltage $V$ remains constant,the new current $I'$ will be $I' = V / (2R) = 1/2 * (V/R) = I/2$.
Therefore,the current becomes half of its original value.

(C) According to Ohm's Law,$V = IR$,which implies $R = V/I$.
Resistance $(R)$ is a property of the conductor that depends on its material,length,cross-sectional area,and temperature.
It does not depend on the current $(I)$ flowing through the circuit or the potential difference $(V)$ applied across it.
Therefore,if the current through the circuit is doubled,the resistance will remain unchanged.

(A) During their motion through a conductor,electrons collide with other electrons and the positive ions of the conductor lattice.
These collisions hinder the flow of electrons,creating an opposition to the electric current.
This opposition to the flow of charge is defined as electrical resistance.

(N/A) The resistivity of a material is defined as the resistance offered by a conductor of that material having a unit length $(1 \ m)$ and a unit cross-sectional area $(1 \ m^2)$.
Mathematically,it is expressed as $\rho = \frac{RA}{l}$,where $R$ is the resistance,$A$ is the cross-sectional area,and $l$ is the length of the conductor.
Its $SI$ unit is $\Omega \cdot m$ (ohm-meter).

(A) The formula for resistivity is $\rho = \frac{R \cdot A}{L}$.
Given:
Resistance $(R)$ = $20\, \Omega$
Area of cross-section $(A)$ = $2\, cm^{2}$
Length $(L)$ = $10\, cm$
Substituting the values into the formula:
$\rho = \frac{20 \times 2}{10} = \frac{40}{10} = 4\, \Omega \cdot cm$.
Thus,the resistivity is $4\, \Omega \cdot cm$.

(A) We know that the formula for resistance is $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.
Since both bulbs are rated for the same voltage $(V = 220\, V)$,the resistance $R$ is inversely proportional to the power $P$ $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lower power rating will have a higher resistance.
Comparing the two bulbs,the $60\, W$ bulb has a lower power rating than the $100\, W$ bulb.
Thus,the $60\, W$ bulb has higher resistance.

(A) According to Ohm's Law,the relationship between voltage $(V)$,current $(I)$,and resistance $(R)$ is given by the formula: $V = I \times R$.
To find the resistance,we rearrange the formula as: $R = \frac{V}{I}$.
Given values are: Voltage $(V)$ = $220\, V$ and Current $(I)$ = $20\, A$.
Substituting these values into the formula: $R = \frac{220}{20} = 11\, \Omega$.
Therefore,the resistance of the electric arc lamp is $11\, \Omega$.

(N/A) An ammeter is designed to have a very low resistance so that it can measure current without significantly altering the circuit. When connected in parallel,the potential difference across the ammeter is the same as the potential difference across the circuit components. Due to its extremely low resistance,a very large amount of current flows through the ammeter according to Ohm's Law $(I = V/R)$. This excessive current generates significant heat $(H = I^2Rt)$,which causes the internal components of the ammeter to overheat and burn out.

(B) An ideal ammeter is designed to measure the current in a circuit without affecting the current flow.
To ensure that the current remains unchanged,an ideal ammeter must have zero resistance.
If the resistance were not zero,it would cause a voltage drop across the ammeter,thereby altering the total current in the circuit.

(D) An ideal voltmeter is designed to measure the potential difference between two points in a circuit without drawing any current from the circuit.
To ensure that no current flows through the voltmeter,it must have an extremely high resistance.
Therefore,the resistance of an ideal voltmeter is considered to be infinite.

(C) Specific resistance, also known as resistivity $(\rho)$, is an intrinsic property of a material.
It depends only on the nature of the material and the temperature of the conductor.
It does not depend on the physical dimensions of the conductor, such as its length or cross-sectional area.
Therefore, since both wires are made of the same material, their specific resistance will be identical regardless of their thickness.

(B) The resistance $R$ of a conductor is directly proportional to its length $l$,given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity and $A$ is the cross-sectional area.
Since $R \propto l$,if the length $l$ is reduced to half $(l' = l/2)$,the new resistance $R'$ will be $R' = \rho \frac{l/2}{A} = \frac{1}{2} R$.
Therefore,the resistance reduces to half of its original value.

(A) When the temperature of a metallic conductor is increased,the atoms or ions in the metal lattice vibrate with greater amplitude. This leads to more frequent collisions between the free electrons and the vibrating atoms,which increases the opposition to the flow of current. Therefore,the resistance of a conductor increases with an increase in temperature.

(B) Copper and aluminium are used for the transmission of electric current because they possess very low electrical resistivity.
Due to this low resistivity,they act as excellent conductors of electricity,allowing current to flow with minimal energy loss in the form of heat.
Additionally,they are relatively inexpensive and possess good mechanical properties,making them suitable for long-distance power transmission.

(D) voltmeter is used to measure the potential difference between two points in a circuit. To ensure that it does not draw any current from the circuit (which would alter the potential difference being measured),it must have a very high resistance. For an ideal voltmeter,the resistance is considered to be infinite.

(A) Iron is a better conductor. The reason is that electrical conductivity is inversely proportional to resistivity. Since the resistivity of iron $(10 \times 10^{-8} \; \Omega m)$ is significantly lower than that of mercury $(94 \times 10^{-8} \; \Omega m)$,iron allows electric current to flow more easily.
$(b)$ Material $A$ will behave as an insulator. Insulators are materials that offer very high resistance to the flow of electric current,which corresponds to a very high resistivity. Since the resistivity of material $A$ $(10^{10}-10^{14} \; \Omega m)$ is extremely high compared to material $B$,it acts as an insulator.

(A) The thick wire will have lower resistance.
According to the formula for resistance,$R = \rho \frac{L}{A}$,where $R$ is resistance,$\rho$ is resistivity,$L$ is length,and $A$ is the area of cross-section.
Since both wires are made of the same material,their resistivity $(\rho)$ is the same.
For a given length,resistance is inversely proportional to the area of cross-section $(R \propto \frac{1}{A})$.
$A$ thick wire has a larger area of cross-section compared to a thin wire.
Therefore,the thick wire offers less opposition to the flow of current,resulting in lower resistance.

(N/A) $Ohm's$ law is not obeyed under the following conditions:
$(i)$ The relationship between potential difference $(V)$ and current $(I)$ is non-linear.
$(ii)$ For a given value of $V$,there is more than one value of $I$ (i.e.,the relation is non-unique).
$(iii)$ The relationship between $V$ and $I$ depends on the sign of $V$ (i.e.,if the direction of current is reversed,the magnitude of current for the same potential difference changes).

(N/A) Alloys like Manganin or Constantan are used to make standard resistors because they have a very low temperature coefficient of resistance. This means that there is a negligible change in the resistance of an alloy with a change in temperature,ensuring that the resistance value remains stable even if the operating temperature fluctuates.

(N/A) Resistance is the property of a conductor to oppose the flow of electric charges through it.
When a potential difference is applied across a conductor,free electrons begin to move,constituting an electric current.
During their motion,these electrons collide with other electrons and atoms of the conductor.
These collisions hinder the flow of electrons,thereby opposing the current.
This opposition to the flow of current is known as resistance.
The $SI$ unit of resistance is the $\text{ohm}$ $(\Omega)$.

(N/A) Conductors are materials that possess a large number of free electrons $(\simeq 10^{28} \text{ m}^{-3})$ and exhibit very low resistivity $(\simeq 10^{-8} \Omega \text{ m})$. The resistivity of an ideal conductor is zero,and it increases with a rise in temperature in metals.
Insulators are substances that have practically no free electrons and possess very high resistivity $(\simeq 10^{16} \Omega \text{ m})$. The resistivity of an ideal insulator is infinity and decreases with a rise in temperature. Mica,rubber,glass,and porcelain are some examples of insulators.

(0.2 A, 2.2 A) According to Ohm's law,the current $I$ is given by the formula $I = \frac{V}{R}$.
For the electric bulb:
Given,$V = 220 \ V$ and $R = 1100 \ \Omega$.
$I = \frac{220}{1100} = 0.2 \ A$.
For the electric heater:
Given,$V = 220 \ V$ and $R = 100 \ \Omega$.
$I = \frac{220}{100} = 2.2 \ A$.