Electric Current Questions in English

(N/A) When conductors are subjected to an electric field,the electric charges within them experience a force.
If these charges are free to move,they will constitute an electric current.
In atoms and molecules,electrons are bound to the nucleus by Coulombic forces. However,in bulk matter,molecules are packed so closely that the outer electrons are no longer strictly attached to a single nucleus.
In metals,some electrons are practically free to move randomly throughout the bulk material. When an external electric field is applied,these free electrons experience a net force and drift,resulting in an electric current.
In solid conductors,current is formed due to the motion of electrons.
In liquids (electrolytes),current is produced due to the motion of positive and negative ions in mutually opposite directions.
In gases,current is produced due to the motion of ions formed by ionization.

(N/A) Consider a metallic cylinder of radius $R$ as shown in the figure.
We consider two thin circular discs which have $+Q$ and $-Q$ charges distributed over them. If we attach these two discs to the two flat surfaces of the cylinder,an electric field $E$ is established,and a current is produced which is directed from the positive to the negative charge.
Due to this electric field in the cylinder,electrons will be accelerated toward the $+Q$ charge.
These electrons,as long as they are moving,will constitute an electric current. In this condition,the current will be formed for a very short time and will then stop as the charges are neutralized.
If we consider a mechanism like a cell or a battery,in which the amount of charge $+Q$ that is neutralized is continuously replenished by an equal amount of $-Q$ charge at the other end,then a steady electric current can be maintained.

(B) In conductors,the atoms have loosely bound electrons in their outermost shells,which are known as free electrons.
When an external electric field is applied across the conductor,these free electrons experience a force and start moving in a specific direction.
The flow of these free electrons constitutes an electric current.
Therefore,current in conductors is formed due to the flow of electrons.

(C) In an electrolyte,the current is formed due to the movement of both positive ions (cations) and negative ions (anions) in opposite directions under the influence of an external electric field. Unlike metallic conductors where current is carried solely by free electrons,electrolytes conduct electricity through the migration of charged ions.

(N/A) In a conductor,free electrons are in a state of continuous random thermal motion due to thermal energy at room temperature.
At any given instant,the number of electrons moving in any direction is equal to the number of electrons moving in the opposite direction.
Mathematically,the average velocity of all free electrons in a conductor in the absence of an external electric field is zero,i.e.,$\vec{v}_{avg} = \frac{1}{N} \sum_{i=1}^{N} \vec{v}_i = 0$.
Since the net flow of charge across any cross-section of the conductor is zero,no net electric current is produced.

(N/A) Inside an electric cell,the electric current flows from the negative electrode (cathode) to the positive electrode (anode). This is due to the chemical energy within the cell that does work to move charges against the electrostatic force,maintaining the potential difference between the terminals.

(B) Inside an electrolyte,the electric current flows from the negative electrode (cathode) to the positive electrode (anode). This is because the positive ions (cations) move towards the cathode and negative ions (anions) move towards the anode,creating a net flow of charge that constitutes the current from the negative terminal to the positive terminal inside the cell.

(N/A) The $SI$ unit of electric charge is the Coulomb $(C)$.
By definition,the current $I$ is the rate of flow of charge $Q$ through a cross-section of a conductor,given by $I = Q/t$.
Therefore,$Q = I \times t$.
When a steady current of $1 \; A$ flows through a conductor for a time interval of $1 \; s$,the quantity of charge that passes through its cross-section is defined as $1 \; C$.
Thus,$1 \; C = 1 \; A \times 1 \; s$.

(N/A) The relationship between electric current $(I)$,charge $(Q)$,and time $(t)$ is given by the formula: $I = \frac{Q}{t}$.
Rearranging this formula to solve for charge,we get: $Q = I \times t$.
By definition,$1 \ Ampere$ is the flow of $1 \ Coulomb$ of charge per second $(1 \ A = 1 \ C/s)$.
Therefore,$1 \ Coulomb$ is defined as the quantity of electric charge that flows through a cross-section of a conductor in $1 \ second$ when a steady current of $1 \ Ampere$ is maintained in the conductor.

(B) The current is given by $i = \alpha_{0} t + \beta t^{2}$.
Substituting the given values,we have $i = 20t + 8t^{2}$.
We know that current $i = \frac{dq}{dt}$,so the charge $q$ is given by the integral $q = \int i \, dt$.
To find the total charge crossed in $15 \, s$,we integrate from $t = 0$ to $t = 15 \, s$:
$q = \int_{0}^{15} (20t + 8t^{2}) \, dt$
$q = \left[ \frac{20t^{2}}{2} + \frac{8t^{3}}{3} \right]_{0}^{15}$
$q = \left[ 10t^{2} + \frac{8}{3}t^{3} \right]_{0}^{15}$
$q = 10(15)^{2} + \frac{8}{3}(15)^{3}$
$q = 10(225) + \frac{8}{3}(3375)$
$q = 2250 + 8(1125)$
$q = 2250 + 9000$
$q = 11250 \, C$.

(C) The correct answer is $(c)$.
In an external circuit,free electrons experience an electrostatic force in the direction opposite to the electric field because they carry a negative charge.
The electric field is always directed from higher potential to lower potential.
Therefore,electrons naturally flow from lower potential to higher potential in an external circuit to reach equilibrium.
However,inside a power source (like a battery),non-electrostatic forces (chemical or magnetic) do work to move electrons from higher potential to lower potential,maintaining the potential difference in the circuit.
Thus,electrons flow from lower to higher potential,except through power sources.

(B) Electric current is defined as the rate of flow of charge through a cross-section.
Although it possesses both magnitude and direction,it does not follow the laws of vector addition (such as the parallelogram law of vector addition).
Instead,it follows the algebraic rules of addition.
Therefore,electric current is classified as a scalar quantity.

(D) The total charge $\Delta q$ flowing through the conductor is equal to the area under the $I-t$ graph.
The area of the triangle is given by:
$\Delta q = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\Delta q = \frac{1}{2} \times 15 \, s \times 10 \, A = 75 \, C$
The average current $I_{\text{avg}}$ is defined as the total charge divided by the total time interval:
$I_{\text{avg}} = \frac{\Delta q}{\Delta t} = \frac{75 \, C}{15 \, s} = 5 \, A$
Therefore,the correct option is $D$.

(D) The charge is given by $Q(t) = \alpha t - \beta t^2 + \gamma t^3$.
The current $i$ is the rate of change of charge with respect to time: $i = \frac{dQ}{dt} = \alpha - 2\beta t + 3\gamma t^2$.
To find the minimum current,we find the derivative of current with respect to time and set it to zero: $\frac{di}{dt} = -2\beta + 6\gamma t = 0$.
Solving for $t$,we get $t = \frac{2\beta}{6\gamma} = \frac{\beta}{3\gamma}$.
Now,substitute $t = \frac{\beta}{3\gamma}$ back into the expression for current:
$i_{min} = \alpha - 2\beta \left(\frac{\beta}{3\gamma}\right) + 3\gamma \left(\frac{\beta}{3\gamma}\right)^2$
$i_{min} = \alpha - \frac{2\beta^2}{3\gamma} + 3\gamma \left(\frac{\beta^2}{9\gamma^2}\right)$
$i_{min} = \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma}$
$i_{min} = \alpha - \frac{\beta^2}{3\gamma}$.

(B) Given that the current $I$ varies with time $t$ as $I = I_0 + \beta t$.
Here,$I_0 = 20 \ A$ and $\beta = 3 \ A/s$.
So,$I = 20 + 3t$.
We know that current $I = \frac{dq}{dt}$,which implies $dq = I \ dt$.
To find the total charge $q$ that crosses a section in time $t = 0$ to $t = 20 \ s$,we integrate the expression:
$q = \int_{0}^{20} I \ dt = \int_{0}^{20} (20 + 3t) \ dt$
$q = \left[ 20t + \frac{3t^2}{2} \right]_{0}^{20}$
$q = \left( 20(20) + \frac{3(20)^2}{2} \right) - 0$
$q = 400 + \frac{3 \times 400}{2} = 400 + 600 = 1000 \ C$.

(A) The electric charge $q$ is related to current $I$ by the relation $q = \int I \ dt$.
Given $I = 3t^2 + 4t^3$,we integrate from $t = 1 \ s$ to $t = 2 \ s$:
$q = \int_{1}^{2} (3t^2 + 4t^3) \ dt$
$q = [t^3 + t^4]_{1}^{2}$
$q = (2^3 + 2^4) - (1^3 + 1^4)$
$q = (8 + 16) - (1 + 1)$
$q = 24 - 2 = 22 \ C$.

(D) The charge $q$ flowing through a wire is given by the integral of current $I$ with respect to time $t$: $q = \int_{t_1}^{t_2} I(t) \, dt$.
Given $I(t) = 0.02t + 0.01$,we integrate from $t = 1 \ s$ to $t = 2 \ s$:
$q = \int_{1}^{2} (0.02t + 0.01) \, dt$
$q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2}$
$q = \left[ 0.01t^2 + 0.01t \right]_{1}^{2}$
Substituting the limits:
$q = (0.01(2)^2 + 0.01(2)) - (0.01(1)^2 + 0.01(1))$
$q = (0.04 + 0.02) - (0.01 + 0.01)$
$q = 0.06 - 0.02 = 0.04 \ C$.

(C) The correct option is $C$.
When a steady current flows through a metallic conductor of non-uniform cross-section,the current $I$ remains constant throughout the conductor due to the principle of conservation of charge.
From the relation for drift velocity,$v_d = \frac{I}{neA}$,since $n$,$e$,and $I$ are constant,$v_d$ depends on the cross-sectional area $A$.
From the relation for current density,$J = \frac{I}{A}$,since $I$ is constant,$J$ depends on $A$.
From the relation for electric field,$E = \rho J = \frac{\rho I}{A}$,since $\rho$ and $I$ are constant,$E$ depends on $A$.
Therefore,only the current $I$ remains constant along the conductor.

(C) The correct answer is current only.
According to the principle of continuity for steady current,the current $I$ flowing through any cross-section of a conductor remains constant,regardless of the area of the cross-section.
Since $I = nAev_d$,where $n$ is the number density of electrons,$A$ is the area of cross-section,$e$ is the charge of an electron,and $v_d$ is the drift speed,if the area $A$ changes,the drift speed $v_d$ must change to keep the current $I$ constant.
Similarly,the electric field $E$ depends on the current density $J = I/A$,so it also varies with the cross-sectional area.

(A) The total charge $Q$ flowing through the cross-section of the conductor is equal to the area under the $I-t$ graph.
From the graph,for $0 \leq t \leq 10 \ s$,the current increases linearly from $100 \ mA$ to $300 \ mA$. The area is a trapezoid: $A_1 = \frac{(100 + 300) \times 10^{-3} \ A}{2} \times 10 \ s = 2 \ C$.
For $10 \ s \leq t \leq 20 \ s$,the current is constant at $300 \ mA$. The area is a rectangle: $A_2 = 300 \times 10^{-3} \ A \times (20 - 10) \ s = 3 \ C$.
Total charge $Q = A_1 + A_2 = 2 \ C + 3 \ C = 5 \ C$.
Using the quantization of charge,$Q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron.
$n = \frac{Q}{e} = \frac{5}{1.6 \times 10^{-19}} = 3.125 \times 10^{19}$.

(B) Given: Current $I = 20 \text{ A}$.
Time $t = 1 \text{ hour } 30 \text{ minutes} = 90 \text{ minutes} = 90 \times 60 \text{ seconds} = 5400 \text{ seconds}$.
The formula for charge transferred is $Q = I \times t$.
Substituting the values: $Q = 20 \text{ A} \times 5400 \text{ s} = 108000 \text{ C}$.
In scientific notation,$Q = 1.08 \times 10^{5} \text{ C}$ or $10.8 \times 10^{4} \text{ C}$.

(A) The current '$I$' flowing through a conductor is defined as the rate of flow of charge with respect to time,given by the derivative of charge '$Q$' with respect to time '$t$':
$I = \frac{dQ}{dt}$
Given the equation for charge: $Q = 3t^2 + t$
Differentiating '$Q$' with respect to '$t$':
$I = \frac{d}{dt}(3t^2 + t) = 6t + 1$
To find the current at time $t = 3 \ s$,substitute '$t = 3$' into the expression for '$I$':
$I = 6(3) + 1 = 18 + 1 = 19 \ A$
Therefore,the current in the conductor at $t = 3 \ s$ is $19 \ A$.

(D) In a metallic conductor,the electric current $I$ is defined as the rate of flow of charge through a cross-section.
For a steady current flowing through a conductor of non-uniform cross-section,the current $I$ remains constant throughout the conductor because charge cannot accumulate at any point.
According to the equation of continuity,$I = nAev_d$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$e$ is the electronic charge,and $v_d$ is the drift velocity.
Since $I$ is constant,if the area $A$ changes,both the current density $J = I/A$ and the drift velocity $v_d = I/(nAe)$ must change to maintain the constant current.
Therefore,only the electric current $I$ remains constant.

(B) Given that,electric current,$I = 9 \,A$.
Let $n$ be the number of electrons passing through a conductor in time $t$.
We know that,$I = \frac{q}{t}$ and $q = ne$.
Therefore,$I = \frac{ne}{t}$.
Rearranging for the number of electrons per second,we get $\frac{n}{t} = \frac{I}{e}$.
Substituting the values,where $e = 1.6 \times 10^{-19} \,C$:
$\frac{n}{t} = \frac{9}{1.6 \times 10^{-19}} = 5.625 \times 10^{19} \approx 5.6 \times 10^{19}$ electrons per second.

(D) The relationship between current $i$ and time $t$ is given by $i = 12t + 9t^2$.
We know that charge $q$ is the integral of current with respect to time: $q = \int_{t_1}^{t_2} i dt$.
Given $t_1 = 2 \,s$ and $t_2 = 10 \,s$,we have:
$q = \int_{2}^{10} (12t + 9t^2) dt$
$q = [12 \cdot \frac{t^2}{2} + 9 \cdot \frac{t^3}{3}]_{2}^{10}$
$q = [6t^2 + 3t^3]_{2}^{10}$
Now,substitute the limits:
$q = (6(10)^2 + 3(10)^3) - (6(2)^2 + 3(2)^3)$
$q = (600 + 3000) - (6 \cdot 4 + 3 \cdot 8)$
$q = 3600 - (24 + 24)$
$q = 3600 - 48 = 3552 \,C$.

(A) The current $I$ is defined as the rate of flow of charge, given by $I = \frac{Q}{t}$.
First, calculate the total charge $Q$ flowing through the wire.
The total number of electrons $N = n \times N_A$, where $n = 0.1 \text{ mol}$ and $N_A = 6 \times 10^{23} \text{ mol}^{-1}$.
$N = 0.1 \times 6 \times 10^{23} = 6 \times 10^{22} \text{ electrons}$.
The total charge $Q = N \times e$, where $e = 1.6 \times 10^{-19} \text{ C}$.
$Q = 6 \times 10^{22} \times 1.6 \times 10^{-19} = 9.6 \times 10^3 \text{ C} = 9600 \text{ C}$.
The time $t = 40 \text{ min} = 40 \times 60 \text{ s} = 2400 \text{ s}$.
Now, calculate the current $I = \frac{9600 \text{ C}}{2400 \text{ s}} = 4 \text{ A}$.

(A) The current is given by $I = I_{0} e^{-\lambda t}$.
We know that current $I$ is the rate of flow of charge,so $I = \frac{dQ}{dt}$.
Therefore,$dQ = I dt = I_{0} e^{-\lambda t} dt$.
To find the total charge $Q$ over the entire pulse period,we integrate from $t = 0$ to $t = \infty$:
$Q = \int_{0}^{\infty} I_{0} e^{-\lambda t} dt$
$Q = I_{0} \left[ \frac{e^{-\lambda t}}{-\lambda} \right]_{0}^{\infty}$
$Q = \frac{I_{0}}{-\lambda} [e^{-\infty} - e^{0}]$
Since $e^{-\infty} = 0$ and $e^{0} = 1$,we get:
$Q = \frac{I_{0}}{-\lambda} [0 - 1] = \frac{I_{0}}{\lambda}$.

(A) The relationship between current $I$ and charge $Q$ is given by $I = \frac{dQ}{dt}$,which implies $Q = \int I \ dt$.
Given $I = 3t^2 + 2t + 5$,we integrate this expression with respect to time from $t = 0$ to $t = 2 \ s$:
$Q = \int_{0}^{2} (3t^2 + 2t + 5) \ dt$
$Q = [t^3 + t^2 + 5t]_{0}^{2}$
$Q = (2^3 + 2^2 + 5(2)) - (0^3 + 0^2 + 5(0))$
$Q = (8 + 4 + 10) - 0$
$Q = 22 \ C$.