Question 13 Marks
Obtain essential expression for balanced position of Wheatstone Bridge.###Explain the principle of Wheatstone Bridge. How can we find the resistance of any resistor using this principle?
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Question 23 Marks
State Kirchhoff's laws and explain them by giving examples.###Explain that first law of Kirchhoff is equivalent to law of conservation of charge and the second law is a descriptive form of Ohm's law.###For a current carrying conductor, write Kirchhoff's laws of charge and energy conservation.
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Question 33 Marks
What is an electric cell? Describe its construction. How many types of electric cells are there? By defining electromotive force of a cell, establish an expression for internal resistance of cell in any circuit and state factors on which internal resistance depends.
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Question 43 Marks
(i) What is meant by electric power? Prove that $P=\frac{V^2}{R}$ or $P=I^2 R$. Define one-watt electric power.(ii) Define electric energy. Prove that $P = I ^2 R \Delta t$
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Question 53 Marks
Establish an expression for mobility by giving definition of mobility. Establish relation between mobility of free electrons and electric current.
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Question 63 Marks
Comment on the following :(i) Electric conductivity.(ii) Dependence of resistivity and resistance on temperature.
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Question 73 Marks
Explain Ohm's law. State factors on which the resistance of any conductor depends. Define specific resistance and state its relation with the conductivity.
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Question 83 Marks
Prove that the magnitude of electric current in a circuit is proportional to the drift velocity of free electrons. Verify Ohm's law with the help of drift velocity.
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Question 93 Marks
Establish a relation between resistance and resistivity of any substance. Explain in detail, the dependence of resistivity on the temperature.
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Question 103 Marks
Obtain the Ohm's law $j=\sigma$ E. Using this relation, establish the relation $V=I R$. What are ohmic and non-ohmic devices? Explain by giving examples.
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Question 113 Marks
Establish a relationship between electric current and drift velocity. Draw the direction of electron flowing in a circuit by drawing an electric circuit.
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Question 123 Marks
What is drift velocity? Explain the dependence of drift velocity and electric field on the mobility.
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Question 133 Marks
Define electric current and current density. By giving an example prove that electric current is a scalar quantity but not a vector quantity. Explain electric current in any conductor.
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Question 143 Marks
Write first rule of Kirchhoff (junction rule).By drawing circuit diagram of Wheatstone Bridge, obtain condition for zero deflection in the bridge.
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Question 153 Marks
Electromotive forces of two cells are $E_1$ and $E _2$ and their internal resistances are $r_1$ and $r_2$. Find equivalent electromotive force and equivalent internal resistance on joining them in series.
Answer
In the figure given, series combination of two cells $E_1$ and $E_2$ of internal resistances $r_1$ and $r_2$ are shown. Its end points are joined to an external resistance R. If current flowing in the resistance R is I due to combination of cells, then by Ohm's law $V = IR$
The magnitude of potential difference across the end points $a$ and $b$ of the combination will be equal to the sum of the terminal voltage of both cells
$V _{a b}=\left( E _1- I r_1\right)+\left( E _2- I r_2\right)$
Terminal voltage will be equal to developed potential difference across the ends of the resistance R. Therefore,
$E _1- I r_1+ E _2- Ir _2= IR$
$E _1+ E _2= IR + I r_1+ I r_2$
$E _1+ E _2= I \left( R +r_1+r_2\right)$
$I =\frac{ E _1+ E _2}{ R +r_1+r_2}=\frac{ E _o}{ R +r_o}$
Here $E _0= E _1+ E _2$ the magnitude of total electromotive force of the cells and $r_0=r_1+r_2$ the total internal resistance of the cells. It is clear from this
(i) The total magnitude of electromotive force in the combination is equal to the sum of the electromotive forces of the combining cells.
(ii) If n cells of same electromotive force and internal resistance are joined in the combination, then the magnitude of current flowing in the external resistance R is
$I =\frac{n E }{ R +n r}=\frac{ E }{r+\frac{ R }{n}}$
which means that the magnitude of current flowing in the external resistance R is greater than the current obtained from a cell.
(iii) If polarity of one battery in the combination is reversed, then the magnitude of electromotive force obtained is either $\left(E_1-E_2\right)$ or $\left(E_2-E_1\right)$.
View full question & answer→In the figure given, series combination of two cells $E_1$ and $E_2$ of internal resistances $r_1$ and $r_2$ are shown. Its end points are joined to an external resistance R. If current flowing in the resistance R is I due to combination of cells, then by Ohm's law $V = IR$
The magnitude of potential difference across the end points $a$ and $b$ of the combination will be equal to the sum of the terminal voltage of both cells
$V _{a b}=\left( E _1- I r_1\right)+\left( E _2- I r_2\right)$
Terminal voltage will be equal to developed potential difference across the ends of the resistance R. Therefore,
$E _1- I r_1+ E _2- Ir _2= IR$
$E _1+ E _2= IR + I r_1+ I r_2$
$E _1+ E _2= I \left( R +r_1+r_2\right)$
$I =\frac{ E _1+ E _2}{ R +r_1+r_2}=\frac{ E _o}{ R +r_o}$
Here $E _0= E _1+ E _2$ the magnitude of total electromotive force of the cells and $r_0=r_1+r_2$ the total internal resistance of the cells. It is clear from this
(i) The total magnitude of electromotive force in the combination is equal to the sum of the electromotive forces of the combining cells.
(ii) If n cells of same electromotive force and internal resistance are joined in the combination, then the magnitude of current flowing in the external resistance R is
$I =\frac{n E }{ R +n r}=\frac{ E }{r+\frac{ R }{n}}$
which means that the magnitude of current flowing in the external resistance R is greater than the current obtained from a cell.
(iii) If polarity of one battery in the combination is reversed, then the magnitude of electromotive force obtained is either $\left(E_1-E_2\right)$ or $\left(E_2-E_1\right)$.
Question 163 Marks
Obtain expression $\vec{J}=\sigma \vec{E}$ of Ohm's law on the basis of drift velocity.
Answer
View full question & answer→Let us consider a conductor of length $/$ and whose area is A . Number density per volume is $n$. The number of charge carriers in the conductor will be $n A /$ where $A /$ is the volume of the conductor. If charge on every charge carrier is $e$ and their drift velocity is, then the total charge of charge carriers in the conductor is $\Delta q=n A / e$ which will pass through cross-section of this conductor in time $\Delta t=\frac{l}{v_d}$. The current flowing through the conductor is $I =\frac{\Delta q}{\Delta t}$.
$\Delta q=n Ale$ and $\Delta t=\frac{l}{v_d}$
$I =\frac{\Delta q}{\Delta t}$
$I =\frac{n A l e}{l / v_d}=n A v_{d e}$
We know that current density,
$j=\frac{ I }{ A }=\frac{n A v_d e}{A}$
$j=n v_d$
In vector form, $\vec{j}=-n e \overrightarrow{v_d}$ (charge on electron is $-e$ ) If applied electric field on the conductor is $\vec{E}$ and average time of collision of charge carriers (electrons) is $\vec{\tau}$, then drift velocity.
$v_d=\frac{-e \overrightarrow{ E }}{m} \vec{\tau}$
$\therefore \quad \vec{j}=-n e\left(\frac{-e \overrightarrow{ E }}{m} \vec{\tau}\right)=\left(\frac{n e^2 \vec{\tau}}{m}\right) \overrightarrow{ E }$
$\vec{j}=\sigma \overrightarrow{ E }$
where, $\sigma=\frac{n e^2 \vec{\tau}}{m}$ is called conductivity of the substance. Equation (3) is a form of ohm's law.
$\Delta q=n Ale$ and $\Delta t=\frac{l}{v_d}$
$I =\frac{\Delta q}{\Delta t}$
$I =\frac{n A l e}{l / v_d}=n A v_{d e}$
We know that current density,
$j=\frac{ I }{ A }=\frac{n A v_d e}{A}$
$j=n v_d$
In vector form, $\vec{j}=-n e \overrightarrow{v_d}$ (charge on electron is $-e$ ) If applied electric field on the conductor is $\vec{E}$ and average time of collision of charge carriers (electrons) is $\vec{\tau}$, then drift velocity.
$v_d=\frac{-e \overrightarrow{ E }}{m} \vec{\tau}$
$\therefore \quad \vec{j}=-n e\left(\frac{-e \overrightarrow{ E }}{m} \vec{\tau}\right)=\left(\frac{n e^2 \vec{\tau}}{m}\right) \overrightarrow{ E }$
$\vec{j}=\sigma \overrightarrow{ E }$
where, $\sigma=\frac{n e^2 \vec{\tau}}{m}$ is called conductivity of the substance. Equation (3) is a form of ohm's law.
Question 173 Marks
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Answer
View full question & answer→Given that : $E =8 V, r=0.5 \Omega, R =15.5 \Omega$,$V _t=120 V$
$V _1- Ir - E - IR =0$
$I=\frac{V_1-E}{R+r}=\frac{120-8}{15.5+0.5}$
current (I) $=\frac{112}{16}=7 A$
Terminal voltage, $V = E + I r$
$=8+7 \times 0.5$
$=8+3.5=11.5$ volt
It limits the current taken from the external source. In its absence, current will increase abruptly which will cause harm to the components.
$V _1- Ir - E - IR =0$
$I=\frac{V_1-E}{R+r}=\frac{120-8}{15.5+0.5}$
current (I) $=\frac{112}{16}=7 A$
Terminal voltage, $V = E + I r$
$=8+7 \times 0.5$
$=8+3.5=11.5$ volt
It limits the current taken from the external source. In its absence, current will increase abruptly which will cause harm to the components.
Question 183 Marks
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27${ }^{\circ} C$ ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times 10^{-4}{ }^{\circ} C ^{-1}$.
Answer
View full question & answer→Given that :
Magnitude of supply voltage, V $=230 V$
Initial current, $I _1=3.2 A$
Room temperature $T _1=27^{\circ} C$
Constant current, $I _2=2.8 A$
Constant temperature $T _2=$ ?
Temperature coefficient of resistance
$\alpha=1.7 \times 10^{-4}{ }^{\circ} C ^{-1}$
If $R_1$ and $R_2$ are resistances of the wire at temperatures $T_1$ and $T_2$ respectively, then
Formula $\quad R _1=\frac{ V }{ I _1}=\frac{230}{3.2}=71.875 \Omega$
$R _2=\frac{ V }{ I _2}=\frac{230}{2.8}=82.143 \Omega$
Formula $\quad \alpha=\frac{ R _2- R _1}{ R _1\left(T_2- T _1\right)}$
$T _2- T _1=\frac{ R _2- R _1}{ R _1 \alpha}$
$\Rightarrow \quad T _2-27^{\circ} C =\frac{82.143-71.875}{71.875 \times 1.7 \times 10^{-4}}=\frac{10.268}{122.19 \times 10^{-4}}$
$\Rightarrow \quad T _2-27^{\circ} C =\frac{10.268}{122.19 \times 10^{-4}}=840.35$
$T _2=840.35+27=867.35^{\circ} C$
Magnitude of supply voltage, V $=230 V$
Initial current, $I _1=3.2 A$
Room temperature $T _1=27^{\circ} C$
Constant current, $I _2=2.8 A$
Constant temperature $T _2=$ ?
Temperature coefficient of resistance
$\alpha=1.7 \times 10^{-4}{ }^{\circ} C ^{-1}$
If $R_1$ and $R_2$ are resistances of the wire at temperatures $T_1$ and $T_2$ respectively, then
Formula $\quad R _1=\frac{ V }{ I _1}=\frac{230}{3.2}=71.875 \Omega$
$R _2=\frac{ V }{ I _2}=\frac{230}{2.8}=82.143 \Omega$
Formula $\quad \alpha=\frac{ R _2- R _1}{ R _1\left(T_2- T _1\right)}$
$T _2- T _1=\frac{ R _2- R _1}{ R _1 \alpha}$
$\Rightarrow \quad T _2-27^{\circ} C =\frac{82.143-71.875}{71.875 \times 1.7 \times 10^{-4}}=\frac{10.268}{122.19 \times 10^{-4}}$
$\Rightarrow \quad T _2-27^{\circ} C =\frac{10.268}{122.19 \times 10^{-4}}=840.35$
$T _2=840.35+27=867.35^{\circ} C$
Question 193 Marks
At room temperature (27.0 ${ }^{\circ} C$) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to 117 Ω, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}{ }^{\circ} C ^{-1}$.
Answer
View full question & answer→Given:
$R _1=100 \Omega$
$R _2=117 \Omega$
$\begin{array}{l} T _1=27^{\circ} C ,
\\ T _2=?\end{array}$
$\alpha=1.70 \times 10^{-4}{ }^{\circ} C ^{-1}$
Formula $\quad R _2= R _1\left[1+\alpha\left( T _2- T _1\right)\right]$
$\Rightarrow \quad \frac{ R _2}{ R _1}=1+\alpha\left( T _2- T _1\right)$
$\Rightarrow \quad \frac{ R _2}{ R _1}-1=\alpha\left( T _2- T _1\right)$
$\Rightarrow \quad \frac{ R _2- R _1}{ R _1}=\alpha\left( T _2- T _1\right)$
$T_2-T_1=\frac{R_2-R_1}{R_1 \alpha}$
On putting the values,
$\Rightarrow \quad T_2-27=\frac{117-100}{100 \times 1.70 \times 10^{-4}}=\frac{17 \times 10^4}{170}$
$\Rightarrow \quad T_2-27=10^5=1000$
$\Rightarrow \quad T _2=1000+27=1027^{\circ} C$
$R _1=100 \Omega$
$R _2=117 \Omega$
$\begin{array}{l} T _1=27^{\circ} C ,
\\ T _2=?\end{array}$
$\alpha=1.70 \times 10^{-4}{ }^{\circ} C ^{-1}$
Formula $\quad R _2= R _1\left[1+\alpha\left( T _2- T _1\right)\right]$
$\Rightarrow \quad \frac{ R _2}{ R _1}=1+\alpha\left( T _2- T _1\right)$
$\Rightarrow \quad \frac{ R _2}{ R _1}-1=\alpha\left( T _2- T _1\right)$
$\Rightarrow \quad \frac{ R _2- R _1}{ R _1}=\alpha\left( T _2- T _1\right)$
$T_2-T_1=\frac{R_2-R_1}{R_1 \alpha}$
On putting the values,
$\Rightarrow \quad T_2-27=\frac{117-100}{100 \times 1.70 \times 10^{-4}}=\frac{17 \times 10^4}{170}$
$\Rightarrow \quad T_2-27=10^5=1000$
$\Rightarrow \quad T _2=1000+27=1027^{\circ} C$
Question 203 Marks
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer
View full question & answer→Given:
$E =10 V, r=3 \Omega$,
$I =0.5 A$
R = ? , V= ?
Formula : $\quad I =\frac{ E }{ R +r}$
or$R+r=\frac{E}{I}$
or$R =\frac{ E }{ I }-r$
Putting value $\quad R =\frac{10}{0.5}-3=20-3$
$R =17 \Omega$
Magnitude of terminal voltage,
$V=E-I r=IR$
$=10-0.5 \times 3=8.5 V$
$E =10 V, r=3 \Omega$,
$I =0.5 A$
R = ? , V= ?
Formula : $\quad I =\frac{ E }{ R +r}$
or$R+r=\frac{E}{I}$
or$R =\frac{ E }{ I }-r$
Putting value $\quad R =\frac{10}{0.5}-3=20-3$
$R =17 \Omega$
Magnitude of terminal voltage,
$V=E-I r=IR$
$=10-0.5 \times 3=8.5 V$