Electric Current and Circuit Questions in English

(D) Free electrons are the charge carriers responsible for the flow of electric current in conductors.
Metals like $Copper$,$Silver$,and $Aluminum$ are good conductors of electricity because they possess a large number of free electrons in their atomic structure.
$Plastic$ is an insulator (non-conductor) of electricity.
In insulators,electrons are tightly bound to their respective atoms and are not free to move,which prevents the flow of electric current.
Therefore,$Plastic$ does not contain free electrons.

(A) The electric force between two stationary point charges is determined by Coulomb's law.
According to Coulomb's law,the magnitude of the electrostatic force $(F)$ between two point charges ($q_1$ and $q_2$) separated by a distance $(r)$ is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The formula is given by: $F = k \cdot \frac{|q_1 \cdot q_2|}{r^2}$,where $k$ is Coulomb's constant.

(B) Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor.
Mathematically,it is expressed as $I = Q / t$,where $I$ is the electric current,$Q$ is the net charge,and $t$ is the time taken.
The $SI$ unit of electric current is the Ampere $(A)$.

(D) The $SI$ unit of electric current is the $Ampere$ $(A)$.
Electric current is defined as the rate of flow of electric charge through a conductor.
One $Ampere$ is equivalent to one $Coulomb$ of charge flowing per second $(1 \ A = 1 \ C/s)$.

(A) Electric current $(I)$ is defined as the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,$I = Q / t$.
The $SI$ unit of charge $(Q)$ is the coulomb $(C)$ and the $SI$ unit of time $(t)$ is the second $(s)$.
Therefore,the unit of electric current is $Coulomb/second$ $(C/s)$,which is defined as an ampere $(A)$.

(D) The prefix $m$ stands for milli,which represents a factor of $10^{-3}$.
Therefore,$1\, mA = 1 \times 10^{-3}\, A$.
Thus,$1\, mA = 10^{-3}\, A$.

(D) The prefix $\mu$ (micro) represents a factor of $10^{-6}$.
Therefore,$1 \, \mu A$ is equal to $10^{-6} \, A$.

(D) We know that $1 \text{ milliampere } (mA) = 10^{-3} \text{ amperes } (A)$.
We also know that $1 \text{ microampere } (\mu A) = 10^{-6} \text{ amperes } (A)$.
To find the value of $1 \text{ mA}$ in terms of $\mu A$,we set up the equation:
$1 \text{ mA} = x \times 1 \mu A$
$10^{-3} \text{ A} = x \times 10^{-6} \text{ A}$
$x = \frac{10^{-3}}{10^{-6}}$
$x = 10^{-3 - (-6)} = 10^{-3 + 6} = 10^{3}$
Therefore,$1 \text{ mA} = 10^{3} \mu A$.

(B) The formula for electric current is $I = Q / t$,where $I$ is current,$Q$ is total charge,and $t$ is time.
Given $I = 1 \text{ A}$ and $t = 1 \text{ s}$,the total charge $Q = I \times t = 1 \text{ A} \times 1 \text{ s} = 1 \text{ C}$.
The quantization of charge is given by $Q = n \times e$,where $n$ is the number of electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \text{ C})$.
To find $n$,we rearrange the formula: $n = Q / e$.
Substituting the values: $n = 1 \text{ C} / (1.6 \times 10^{-19} \text{ C}) = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons pass through the conductor per second.

(C) An $Ammeter$ is an instrument used to measure the electric current in a circuit. It is always connected in series with the component through which the current is to be measured. $A$ $Voltmeter$ measures potential difference,a $Galvanometer$ detects small amounts of current,and a $Wattmeter$ measures electric power.

(C) By convention,the direction of electric current is taken as the direction of flow of positive charges.
Since electrons are negatively charged particles,they flow from the negative terminal to the positive terminal of a source.
Therefore,the direction of conventional current is considered to be opposite to the direction of the flow of electrons.

(A) The simple battery,known as the voltaic pile,was invented by the Italian physicist Alessandro Volta in $1800$.
It was the first electrical battery that could provide a continuous current to a circuit.
Therefore,the correct option is $A$.

(B) In a Voltaic cell (or any electrochemical cell),the chemical reactions occur at the electrodes.
In the external circuit,electrons flow from the anode (negative electrode) to the cathode (positive electrode).
This flow of electrons constitutes the electric current,which is conventionally defined as flowing from the positive terminal to the negative terminal (opposite to the direction of electron flow).
Therefore,the correct direction of electron flow is from the negative electrode to the positive electrode.

(A) In an electric circuit,the symbol for an electric cell consists of two parallel vertical lines of unequal length. The longer line represents the positive terminal $(+)$,and the shorter,thicker line represents the negative terminal $(-)$.
Option $A$ correctly represents this symbol.

(C) The $SI$ unit of electric charge is the coulomb $(C)$.
$1$ coulomb is defined as the quantity of charge that passes through a point in a circuit when a current of $1$ ampere flows for $1$ second.
Therefore,the correct option is $C$.

(C) Given:
Electric current $(I)$ = $0.5\;A$
Time $(t)$ = $1\;hour = 60 \times 60\;seconds = 3600\;s$
Formula:
Electric charge $(Q)$ = $I \times t$
Calculation:
$Q = 0.5\;A \times 3600\;s$
$Q = 1800\;C$
Therefore, the total electric charge passing through the bulb is $1800\;C$.

(C) The formula for electric current $(I)$ is given by $I = \frac{Q}{t}$, where $Q$ is the charge and $t$ is the time in seconds.
Given: Charge $(Q) = 1800\,C$, Time $(t) = 1\,hour = 60 \times 60\,s = 3600\,s$.
Substituting the values into the formula: $I = \frac{1800}{3600} = 0.5\,A$.
Therefore, the current flowing through the bulb is $0.5\,A$.

(C) The current $I$ drawn by an appliance is given by the formula $P = V \times I$,which implies $I = P / V$.
We calculate the current for each option:
$A) I = 40 \ W / 110 \ V \approx 0.36 \ A$
$B) I = 75 \ W / 220 \ V \approx 0.34 \ A$
$C) I = 60 \ W / 110 \ V \approx 0.54 \ A$
$D) I = 100 \ W / 220 \ V \approx 0.45 \ A$
Comparing these values,the appliance with the rating $110 \ V / 60 \ W$ draws the maximum current of approximately $0.54 \ A$.

(A) battery is a device that converts chemical energy into electrical energy.
It produces a steady flow of electrons in a single direction,which is known as Direct Current $(DC)$.
Unlike Alternating Current $(AC)$,which periodically reverses its direction,the current from a battery remains constant in its polarity and direction over time.
Therefore,the correct option is $A$.

(B) $Galvanometer$ is an electromechanical instrument used for detecting and indicating an electric current.
It works on the principle that a magnetic field exerts a torque on a current-carrying coil placed within it.
When a small current flows through the coil,it deflects,indicating the presence and direction of the current.
In contrast,a $Voltmeter$ measures potential difference,a $Fuse$ is a safety device,and a $Battery$ is a source of electric energy.

(D) Direct Current $(DC)$ is a type of electric current that flows in only one direction.
Unlike Alternating Current $(AC)$,which periodically reverses its direction and magnitude,$DC$ maintains a constant polarity and magnitude over time.
Frequency is defined as the number of cycles per second of an alternating waveform.
Since $DC$ voltage does not oscillate or change its direction,it does not complete any cycles.
Therefore,the frequency of $DC$ voltage is $0 \ Hz$.

(C) Direct current $(DC)$ is a unidirectional flow of electric charge.
Most electronic devices that contain batteries or sensitive electronic circuits,such as a cellphone,operate on $DC$.
Appliances like refrigerators,tubelights,and electric fans typically run on alternating current $(AC)$ supplied by the power grid.