Electric Current and Circuit Questions in English

(A) An electric circuit is a continuous and closed path of an electric current.
It consists of electric devices,switching devices,and a source of electricity,all connected by conducting wires.

(A) The unit of electric current is the ampere $(A)$.
$1$ ampere is defined as the flow of $1$ coulomb $(C)$ of electric charge through a cross-section of a conductor in $1$ second $(s)$.
Mathematically,$I = Q / t$,where $I$ is current in amperes,$Q$ is charge in coulombs,and $t$ is time in seconds.

(N/A) One electron possesses a charge of $1.6 \times 10^{-19} \, C$,which means $1.6 \times 10^{-19} \, C$ of charge is contained in $1$ electron.
Using the formula $Q = ne$,where $Q = 1 \, C$ and $e = 1.6 \times 10^{-19} \, C$:
$n = \frac{Q}{e} = \frac{1}{1.6 \times 10^{-19}} = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons constitute one coulomb of charge.

(D) The formula for electric current is $I = Q / t$,where $I$ is the current,$Q$ is the total charge,and $t$ is the time.
Given $I = 1\, A$ and $t = 16\, s$,the total charge $Q = I \times t = 1\, A \times 16\, s = 16\, C$.
The charge of a single electron is $e = 1.6 \times 10^{-19}\, C$.
The number of electrons $n$ is given by the formula $Q = n \times e$,which implies $n = Q / e$.
Substituting the values: $n = 16 / (1.6 \times 10^{-19}) = 10 / 10^{-19} = 10^{20}$.
Therefore,the number of electrons passing through the filament is $10^{20}$.

(D) To determine the correct circuit,we must follow these rules for connecting electrical components:
$1$. An ammeter $(A)$ must always be connected in series with the circuit component (resistor $R$) to measure the current flowing through it. Its positive terminal should be connected towards the positive terminal of the battery.
$2$. $A$ voltmeter $(V)$ must always be connected in parallel across the component (resistor $R$) whose potential difference is to be measured. Its positive terminal should be connected towards the positive terminal of the battery.
Evaluating the circuits:
- Circuit $(i)$: The voltmeter is connected in series,which is incorrect.
- Circuit $(ii)$: The ammeter is connected in parallel,which is incorrect.
- Circuit $(iii)$: The ammeter is connected in parallel,which is incorrect.
- Circuit $(iv)$: The ammeter is connected in series with the resistor,and the voltmeter is connected in parallel across the resistor. Both are connected with the correct polarity. Thus,this is the correctly connected circuit.

(A) To obtain the maximum potential difference from a combination of cells in series,the positive terminal of one cell must be connected to the negative terminal of the next cell.
In this arrangement,the potentials of the individual cells add up $(V_{total} = V_1 + V_2 + V_3 + ...)$.
Looking at the provided diagrams:
- In diagram $(i)$,the cells are connected such that the positive terminal of one is connected to the negative terminal of the next,which is the correct series arrangement.
- In the other diagrams,some cells are connected in reverse polarity,which would subtract from the total potential.
Therefore,$(i)$ is the correct representation.

(A) The quantization of electric charge is given by the formula $q = ne$,where $q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge.
Given $q = 1 \text{ C}$ and $e = 1.6 \times 10^{-19} \text{ C}$.
To find the number of electrons $(n)$,we rearrange the formula: $n = q / e$.
Substituting the values: $n = 1 / (1.6 \times 10^{-19}) = 10^{19} / 1.6 = 6.25 \times 10^{18}$.
Therefore,there are approximately $6.25 \times 10^{18}$ electrons in $1 \text{ C}$ of charge.

(A) The total charge $Q$ is given as $1 \text{ C}$.
The charge on a single electron $e$ is $1.6 \times 10^{-19} \text{ C}$.
Using the quantization of charge formula, $Q = ne$, where $n$ is the number of electrons.
Therefore, $n = Q / e$.
$n = 1 \text{ C} / (1.6 \times 10^{-19} \text{ C})$.
$n = 1 / 1.6 \times 10^{19}$.
$n = 0.625 \times 10^{19}$.
$n = 6.25 \times 10^{18}$ electrons.

(N/A) The instrument used to measure electric current in a circuit is called an $Ammeter$.
$(b)$ An $Ammeter$ is always connected in $series$ with the component or circuit through which the current is to be measured. This ensures that the total current flowing through the circuit passes through the $Ammeter$.

(N/A) The direction of conventional current is defined as the direction in which positive charges would move. Since electrons are negatively charged,they flow in the direction opposite to the conventional current. Therefore,the direction of conventional current is opposite to the direction of flow of electrons.

(B) In metallic conductors,the flow of electric current is primarily due to the movement of free electrons. These electrons are negatively charged particles that move through the crystal lattice of the conductor when an external electric field is applied.

(B) The electric current in a circuit is measured using a device called an $Ammeter$.
An $Ammeter$ is always connected in series with the component through which the current is to be measured.

(A) An ammeter is a device used to measure the electric current flowing through a circuit. To measure the total current passing through the components,it must be connected in series with the circuit elements so that the same current flows through it.

(N/A) Electric current is defined as the rate of flow of electric charge through a conductor. Mathematically,it is the amount of charge $Q$ flowing through a cross-section of a conductor in time $t$,expressed as $I = Q/t$.
The $SI$ unit of electric current is the ampere $(A)$,which is equivalent to one coulomb per second $(1 \ A = 1 \ C/s).$

(A) An ammeter is a device used to measure the electric current in a circuit.
To measure current accurately,it must be connected in series with the circuit components.
If an ammeter had high resistance,it would significantly reduce the total current flowing through the circuit,leading to inaccurate readings.
Therefore,an ideal ammeter is designed to have a very low resistance so that it does not impede the flow of current.

(N/A) An electric circuit is a continuous and closed path along which an electric current flows.
An open circuit is a circuit in which the path is broken or the switch is in the $OFF$ position,preventing the flow of current.
$A$ closed circuit is a circuit in which the path is complete and the switch is in the $ON$ position,allowing the electric current to flow continuously.

(A) Given: Charge $Q = 12 \, C$,time $t = 4 \, s$.
The formula for electric current $I$ is given by $I = \frac{Q}{t}$.
Substituting the values,we get $I = \frac{12 \, C}{4 \, s} = 3 \, A$.
Therefore,the current supplied by the cell is $3 \, A$.

(A) Given: Current $I = 1 \, A$, Time $t = 1 \, s$.
We know that charge $Q = I \times t$.
Therefore, $Q = 1 \, A \times 1 \, s = 1 \, C$.
The charge of a single electron is $e = 1.6 \times 10^{-19} \, C$.
The number of electrons $n$ is given by the formula $n = \frac{Q}{e}$.
Substituting the values: $n = \frac{1 \, C}{1.6 \times 10^{-19} \, C}$.
$n = 0.625 \times 10^{19} = 6.25 \times 10^{18}$ electrons per second.

(N/A)

Voltmeter	Ammeter
$(a)$ It is used to measure the potential difference between two points in a circuit.	$(a)$ It is used to measure the electric current flowing through a circuit.
$(b)$ It is always connected in parallel across the component.	$(b)$ It is always connected in series with the component.
$(c)$ It has a very high resistance to ensure minimal current flows through it.	$(c)$ It has a very low resistance to ensure minimal voltage drop across it.

(N/A) The values are as follows:
$1 \text{ mA (milliampere)} = 10^{-3} \text{ A} = \frac{1}{1000} \text{ A}$
$1 \text{ } \mu\text{A (microampere)} = 10^{-6} \text{ A} = \frac{1}{1000000} \text{ A}$
$(b)$ The symbols for a battery and a rheostat are shown in the image provided.

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

(N/A) The current through a conductor is said to be $1$ ampere if a charge of $1$ coulomb flows through it in $1$ second. Mathematically,it is expressed as $I = Q/t$,where $I = 1 \text{ A}$,$Q = 1 \text{ C}$,and $t = 1 \text{ s}$.

(720 C) The formula for electric charge is $Q = I \times t$.
Given:
Current $(I)$ = $0.2 A$
Time $(t)$ = $1 \text{ hour} = 3600 s$
Calculation:
$Q = 0.2 A \times 3600 s$
$Q = 720 C$
Therefore,the total charge flowing through the bulb is $720 C$.

(A) Given:
Current $(I)$ = $0.5\, A$
Time $(t)$ = $1\, \text{hour} = 60 \times 60 = 3600\, s$
Voltage $(V)$ = $200\, V$
Formula for electric charge is $Q = I \times t$
Substituting the values:
$Q = 0.5\, A \times 3600\, s = 1800\, C$
Therefore, the amount of electric charge flowing through the electric iron is $1800\, C$.

(1440 C) The formula for electric charge is $Q = I \times t$.
Given:
Current $(I)$ = $0.4\, A$
Time $(t)$ = $1\, \text{hour} = 60 \times 60\, s = 3600\, s$.
Substituting the values into the formula:
$Q = 0.4\, A \times 3600\, s$
$Q = 1440\, C$.
Therefore, the total charge flowing through the appliance is $1440\, C$.

(A) Direct Current $(DC)$ flows in a constant direction and maintains a constant magnitude over time.
Since the current does not oscillate or change its direction periodically,it does not complete any cycles per second.
Therefore,the frequency of $DC$ is $0 \ Hz$.

(B) The statement is $False$.
Electric current that flows in a single direction is known as $Direct \ Current$ $(DC)$.
$Alternating \ Current$ $(AC)$ is the current that periodically reverses its direction.

(A) Rubber is an insulator,meaning it does not allow electric current to flow through it. By wearing rubber gloves and rubber shoes,a person creates a high-resistance barrier that prevents electricity from passing through the body to the ground,thereby protecting the individual from electric shocks.

(N/A) Based on standard circuit symbols:
$A$ represents an electric bulb,which is a load in the circuit.
$B$ represents a rheostat (variable resistor),used to control the current in the circuit.
$C$ represents an electric cell,which acts as the source of potential difference.
$D$ represents an ammeter,which is used to measure the electric current flowing through the circuit.

(A) The $SI$ unit of electric current is the ampere $(A)$.
One ampere is defined as the flow of $1$ coulomb of charge per second through a conductor $(1 \text{ A} = 1 \text{ C/s})$.
To calculate the number of electrons $(n)$:
Given: Current $(I)$ = $1 \text{ A}$,Time $(t)$ = $1 \text{ s}$,Charge on an electron $(e)$ = $1.6 \times 10^{-19} \text{ C}$.
Using the formula $Q = I \times t$ and $Q = n \times e$,we get $n = \frac{I \times t}{e}$.
$n = \frac{1 \text{ A} \times 1 \text{ s}}{1.6 \times 10^{-19} \text{ C}} = \frac{1}{1.6} \times 10^{19} = 6.25 \times 10^{18}$.
Thus,$6.25 \times 10^{18}$ electrons flow through the conductor in $1$ second.

(N/A) The direction of electric current is conventionally defined as opposite to the direction of the flow of electrons in a wire.
$(b)$ Given:
Charge $(Q) = 500 \, C$
Time $(t) = 10 \, \text{minutes} = 10 \times 60 \, \text{s} = 600 \, \text{s}$
Using the formula for electric current:
$I = \frac{Q}{t}$
$I = \frac{500 \, C}{600 \, \text{s}}$
$I = 0.833 \, \text{A}$
Therefore,the current in the circuit is approximately $0.833 \, \text{A}$.

(A) Given: Current $I = 0.5\, A$, Time $t = 10\, \text{min} = 10 \times 60 = 600\, \text{s}$.
The formula for electric charge $Q$ is $Q = I \times t$.
Substituting the values: $Q = 0.5\, A \times 600\, \text{s} = 300\, C$.
Therefore, the amount of electric charge that flows through the circuit is $300\, C$.

(N/A) An electric circuit is a continuous and closed path in which an electric current flows.
The circuit diagram consists of a battery of two cells,a resistor,an ammeter,and a closed plug key connected in series. The ammeter is connected in series with the resistor,and the positive terminal of the battery is connected to the positive terminal of the ammeter.

(N/A) The bulb will glow only when the piece of silver is placed between points $A$ and $B$.
Silver is a metal and acts as a good conductor of electricity,which completes the electrical circuit and allows current to flow through the bulb.
In contrast,wood is an insulator (a bad conductor of electricity). When a piece of wood is placed between points $A$ and $B$,it breaks the electrical circuit,preventing the flow of current,and therefore the bulb does not glow.

(A) The $SI$ unit of electric charge is the coulomb $(C)$. It is defined as the amount of charge transported by a constant current of one ampere in one second $(1 \ C = 1 \ A \cdot s)$.

(A) In a metal,electrons in the outermost shells of atoms are loosely bound to the nucleus. These are called free electrons. They can move randomly throughout the entire volume of the metal conductor. However,they cannot escape the metal surface under normal conditions because there is a potential barrier at the surface that keeps them confined within the material.

(B) The definition of electric current $(I)$ is the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,current is defined as the quantity of charge flowing past a point divided by the time $(t)$ taken for that flow.
Formula: $I = Q / t$.
Therefore,the statement that current is charge multiplied by time is incorrect.

(A) cell is a device that converts chemical energy into electrical energy. Inside the cell,chemical reactions occur that create a potential difference between the two terminals. This potential difference drives the flow of electrons (charge) through the connected conducting wire,resulting in an electric current.

(D) 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.
- Ampere $(A)$ is the $SI$ unit of electric current.
- Volt $(V)$ is the $SI$ unit of electric potential difference.
- Watt $(W)$ is the $SI$ unit of power.

(C) The quantization of electric charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge of an electron.
Given: Total charge $Q = 1.6 \ C$ and the charge of one electron $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $n$: $n = Q / e$.
Substituting the values: $n = 1.6 \ C / (1.6 \times 10^{-19} \ C) = 1 / 10^{-19} = 10^{19}$.
Therefore,there are $10^{19}$ electrons in a charge of $1.6 \ C$.

(B) The prefix $\mu$ (micro) represents $10^{-6}$ and the prefix $m$ (milli) represents $10^{-3}$.
To convert $1\, \mu A$ to $mA$,we use the following calculation:
$1\, \mu A = 10^{-6}\, A$
$1\, mA = 10^{-3}\, A$
Therefore,$1\, \mu A = \frac{10^{-6}}{10^{-3}}\, mA = 10^{-6 - (-3)}\, mA = 10^{-3}\, mA$.

(A) Free electrons are the charge carriers responsible for electrical conductivity in materials.
Materials are classified as conductors,insulators,or semiconductors based on the availability of free electrons.
$A$. Copper is a metal and an excellent conductor of electricity because it possesses a large number of free electrons.
$B$. Glass,$C$. Rubber,and $D$. Paper are insulators,meaning they have very few or negligible free electrons,which prevents the flow of electric current.
Therefore,Copper has the highest number of free electrons among the given options.

(B) Electric current $(I)$ is defined as the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,it is expressed as the amount of charge flowing per unit time $(t)$.
Therefore,the formula is $I = Q / t$,where $I$ is current in Amperes,$Q$ is charge in Coulombs,and $t$ is time in seconds.

(D) The formula for electric charge $(Q)$ is given by the product of electric current $(I)$ and time $(t)$: $Q = I \times t$.
Given:
Electric current $(I)$ = $2 \ A$
Time $(t)$ = $1 \ minute = 60 \ seconds$.
Substituting the values into the formula:
$Q = 2 \ A \times 60 \ s = 120 \ C$.
Therefore,the total electric charge passing through the wire is $120 \ C$.

(C) The formula for electric current $(I)$ is given by $I = \frac{Q}{t}$,where $Q$ is the total charge and $t$ is the time.
Since $Q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \text{ C})$,the formula becomes $I = \frac{ne}{t}$.
Given: $I = 4.8 \text{ A}$,$t = 1 \text{ s}$,and $e = 1.6 \times 10^{-19} \text{ C}$.
Rearranging for $n$: $n = \frac{I \times t}{e}$.
Substituting the values: $n = \frac{4.8 \times 1}{1.6 \times 10^{-19}}$.
$n = 3 \times 10^{19}$ electrons.
Therefore,the number of electrons passing through the appliance is $3 \times 10^{19}$.

(A) proton is a subatomic particle found in the nucleus of an atom.
It carries a fundamental positive electric charge of approximately $+1.602 \times 10^{-19} \ C$.
Therefore,the correct option is $A$.

(C) An electron is a subatomic particle that carries a fundamental negative electric charge. By convention,the charge of an electron is defined as $-1.602 \times 10^{-19} \ C$.

(A) The elementary charge,denoted by $e$,is the electric charge carried by a single proton or,equivalently,the magnitude of the negative electric charge carried by a single electron.
It is a fundamental physical constant.
The value of the charge on an electron is approximately $1.602 \times 10^{-19} \ C$.
Therefore,the correct magnitude is $1.6 \times 10^{-19} \ C$.

(B) The quantization of electric charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge of an electron.
Given: $Q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $n$: $n = Q / e$.
Substituting the values: $n = 1 / (1.6 \times 10^{-19})$.
$n = 10^{19} / 1.6$.
$n = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons constitute a total charge of $1 \ C$.

(C) The charge of a single proton is $e = 1.6 \times 10^{-19} \ C$.
To find the total charge $(Q)$ on $n = 100$ protons,we use the formula $Q = n \times e$.
Substituting the values: $Q = 100 \times (1.6 \times 10^{-19} \ C)$.
$Q = 10^2 \times 1.6 \times 10^{-19} \ C$.
$Q = 1.6 \times 10^{-17} \ C$.
Therefore,the total electric charge is $1.6 \times 10^{-17} \ C$.