**1** **What does an electric circuit mean?**

A continuous and closed path of an electric current is called an electric circuit. electric circuit consists of electric devices like switching devices, source of electricity, etc.

**2** **Define the unit of current.**

The unit of electric current is ampere (A).$1$A is defined as the flow of $1 $C of charge through a wire in $1$ s of time.

**3** **Calculate the number of electrons constituting one coulomb of charge.**

We know that charge is quantized i.e., Q = ne where n = number of electrons and e = charge of an electron , Therefore Given that$\\$ $Q=1C,e=1.6*10^{-19}C,n=?$$\\$ On substitution the above values in Q = ne, We get,$\\$ $1C=n*1.6*10^{-19}=\\ \dfrac{1}{1.6*110^{-19}}=\\ 6.25*10^{18}$$\\$ $=6*10^{18}$ electrons$\\$ Therefore, $6*10^{18 } $electrons constitute one coulomb of charge.

**4** **Name a device that helps to maintain a potential difference across a conductor.**

A source of electricity such as cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

**5** **What is meant by saying that the potential difference between two points is $1 V$?**

If $1 J$ of work is required to move a charge of amount $1 C$ from one point to another,then it is said that the potential difference between the two points is $1 V$.

**6** **How much energy is given to each coulomb of charge passing through a $6 V$ battery?**

The energy given to each coulomb of charge is equal to the amount of work done to move it. The amount of work is given by the expression,$\\$ Potential difference =$\dfrac{\text{Work Done}}{\text{Charge}}$$\\$ Work Done = Potential Difference × Charge$\\$ Where,$\\$ Charge =$ 1 C$ $\\$ Potential difference =$ 6 V$ $\\$ Work Done = $6 × 1 = 6 J$ $\\$ Therefore, $6 J$ of energy is given to each coulomb of charge passing through a battery of $6 V.$

**7** **On what factors does the resistance of a conductor depend?**

The resistance of a conductor depends upon the following factors:$\\$ (a) Length of the conductor (l) directly$\\$ (b) Cross-sectional area of the conductor (A) inversely$\\$ (c) Nature of Material of the conductor$\\$ (d) Temperature of the conductor

**8** **Will current flow more easily through a thick wire or a thin wire of the same material,when connected to the same source? Why?**

Resistance of a wire, $R =\rho \dfrac{l}{A}$ $\\$ Where,$\\$ $\rho =$ Resistivity of the material of the wire$\\$ $l =$ Length of the wire$\\$ A = Area of cross-section of the wire$\\$ Resistance is inversely proportional to the area of cross-section of the wire.Thus Thicker the wire, lower is the resistance of the wire and vice-versa. Therefore, current can flow more easily through a thick wire than a thin wire.

**9** **Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value.What change will occur in the current through it?**

The change in the current flowing through the component is given by Ohm’s law as,$\\$ $V = IR$ $\\$ $I=\dfrac{V}{R}$ $\\$ Where,$\\$ R = Resistance of the electrical component.$\\$ V = Potential difference$\\$ I = Current $\\$ The potential difference is reduced to half i.e., $V' = (V/2)$, keeping resistance constant.$\\$ Let the new resistance be $R'= R$ and the new amount of current be $I '$.$\\$ Therefore, from Ohm’s law, we obtain the amount of new current.$\\$ $ I'=\dfrac{V'}{R'}=\dfrac{\dfrac{V}{2}}{R}=\dfrac{1}{2}(\dfrac{V}{R})=\dfrac{I }{2}$ $\\$ Therefore, the amount of current flowing through the electrical component is reduced by half.

**10** **Why are coils of electric toasters and electric irons made of an alloy rather than a puremetal?**

The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures,the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

**11** **Use the data in Table to answer the following $-\\$ Table: Electrical resistivity of some substances at $20°C$$\\$ $\begin{array}{|c|c|}\hline & \text{Material}& \text{Resistivity}(\Omega m)\\ \hline \text{Conductors}& \text{Silver}& 1.60*10^{-8}\\ & \text{Copper}& 1.62*10^{-8}\\ & \text{Aluminium}& 2.63*10^{-8}\\ & \text{Tungsten} & 5.20* 10^{-8}\\ & \text{Nickel}& 6.84*10^{-8}\\ & \text{Iron}& 10.0*10^{-8}\\ & \text{Chromium}& 12.9*10^{-8}\\ & \text{Mercury}& 94.0*10^{-8}\\ & \text{Manganese} & 1.84*10^{-6}\\ \hline \text{Alloys}& \text{Constantan(alloy of Cu and Ni)}&49*10^{-6}\\ & \text{Manganin(alloy of Cu,Mn and Ni)}& 44* 10^{-6}\\ & \text{Nichrome(alloy of Ni,Cr,Mn and Fe)} & 100* 10^{-6}\\ \hline \text{Insulators}& \text{Glass}& 10^{10}-10^{14}\\ &\text{Hard rubber} & 10^{13}-10^{16}\\ & \text{Ebonite} & 10^{15}-10^{17}\\ & \text{Diamond}& 10^{12}-10^{13}\\ & \text{Paper(dry)}& 10^{12}\\ \hline \end{array}$$\\$ (a) Which among iron and mercury is a better conductor?$\\$ (b) Which material is the best conductor?**

(a) Resistivity of iron $= 10.0 * 10^{-8} \Omega m$ $\\$ Resistivity of mercury $= 94.0 * 10^{-8 }\Omega m$ $\\$ Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.$\\$ (b) It can be observed from Table that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.