exemplar-phy-12-current-electricity

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<p><strong>Ans.</strong> (b)</p>
<p>Current per unit area (taken normal to the current), $I/A$, is called current density and is denoted by $j$. The SI units of the current density are A $/ m^{2}$. The current density is also directed along $E$ and is also a vector and the relationship is given by</p>
<p>$j=s E$</p>
<p>The $j$ changes due to electric field produced by charges accumulated on the surface of wire.</p>
<p>Note That as the currrent progresses along the wire, the direction of $j$ (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for this.</p>
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<p><strong>Ques.</strong> Two batteries of emf $\varepsilon_1$ and $\varepsilon_2(\varepsilon_2>\varepsilon_1)$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure.</p>
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/cf4b7305-7993-4fff-53da-93ae0b27e800/public" height="100px">
<div class="fragment">
<p>(a) Two equivalent emf $\varepsilon   _{\text{eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e., $\varepsilon_1<\varepsilon _{\text{eq }}<\varepsilon_2$ <br>
(b) The equivalent emf $\varepsilon   _{\text{eq }}$ is smaller than $\varepsilon_1$ <br>
(c) The $\varepsilon   _{\text{eq }}$ is given by $\varepsilon _{\text{eq }}=\varepsilon_1+\varepsilon_2$ always <br>
(d) $\varepsilon   _{\text{eq }}$ is independent of internal resistances $r_1$ and $r_2$</p>
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<p><strong>Ques.</strong> A resistance $R$ is to be measured using a meter bridge, student chooses the standard resistance $S$ to be $100 \Omega$. He finds the null point at $I_1=2.9 cm$. He is told to attempt to improve the accuracy. Which of the following is a useful way?</p>
<div class="fragment">
<p>(a) He should measure $I_1$ more accurately</p>
<p>(b) He should change $S$ to $1000 \Omega$ and repeat the experiment</p>
<p>(c) He should change $S$ to $3 \Omega$ and repeat the experiment</p>
<p>(d) He should given up hope of a more accurate measurement with a meter bridge</p>
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<p><strong>Thinking Process</strong></p>
<p>Here, the concept of accurate balanced Wheatstone bridge is to be used.</p>
<p><strong>Ans.</strong> (c)</p>
<p>The percentage error in $R$ can be minimised by adjusting the balance point near the middle of the bridge, i.e., when $I_1$ is close to $50 cm$. This requires a suitable choice of $S$.</p>
<p>Since,<br>
$\frac{R}{S}=\frac{R l_1}{R(100-l_1)}=\frac{l_1}{100-l_1}$</p>
<p>Since here, $R: S:: 2.9: 97.1$ imply that the $S$ is nearly 33 times to that of $R$. In orded to make this ratio $1: 1$, it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$.</p>
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<p><strong>Ques.</strong> Two cells of emfs approximately $5 V$ and $10 V$ are to be accurately compared using a potentiometer of length $400 cm$.</p>
<div class="fragment">
<p>(a) The battery that runs the potentiometer should have voltage of $8 V$</p>
<p>(b) The battery of potentiometer can have a voltage of $15 V$ and $R$ adjusted so that the potential drop across the wire slightly exceeds $10 V$</p>
<p>(c) The first portion of $50 cm$ of wire itself should have a potential drop of $10 V$</p>
<p>(d) Potentiometer is usually used for comparing resistances and not voltages</p>
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<p><strong>Thinking Process</strong></p>
<p>The potential drop across wires of potentiometer should be greater than emfs of primary cells.</p>
<p><strong>Ans.</strong> (b)</p>
<p>In a potentiometer experiment, the emf of a cell can be measured, if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as $5 V$ and $10 V$, therefore, the potential drop along the potentiometer wire must be more than $10 V$.</p>
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<p><strong>Ques.</strong> A metal rod of length $10 cm$ and a rectangular cross-section of $1 cm \times \frac{1}{2} cm$ is connected to a battery across opposite faces. The resistance will be</p>
<div class="fragment">
<p>(a) maximum when the battery is connected across $1 cm \times \frac{1}{2} cm$ faces</p>
<p>(b) maximum when the battery is connected across $10 cm \times 1 cm$ faces</p>
<p>(c) maximum when the battery is connected across $10 cm \times \frac{1}{2} cm$ faces</p>
<p>(d) same irrespective of the three faces</p>
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<p><strong>Thinking Process</strong></p>
<p>The resistance of wire depends on its geometry $l$ (length of the rod). Here, the metallic rod behaves as a wire.</p>
<p><strong>Ans.</strong> (a)</p>
<p>The resistance of wire is given by</p>
<p>$R=\rho \frac{l}{A}$</p>
<p>For greater value of $R, l$ must be higher and $A$ should be lower and it is possible only when the battery is connected across $1 cm \times \frac{1}{2} cm$ (area of cross-section $A$ ).</p>
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<p><strong>Ques.</strong> Consider a simple circuit shown in figure stands for a variable resistance $R^{\prime} . R^{\prime}$ can vary from $R_0$ to infinity. $r$ is internal resistance of the battery $(r«R«R_0$).</p>
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/80eeab4d-0440-4333-d7c4-c9062be33d00/public" height="120px">
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<p>(a) Potential drop across $A B$ is nearly constant as $R^{\prime}$ is varied<br>
(b) Current through $R^{\prime}$ is nearly a constant as $R^{\prime}$ is varied<br>
(c) Current $I$ depends sensitively on $R^{\prime}$<br>
(d) $I \geq \frac{V}{r+R}$ always</p>
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<p><strong>Ques.</strong> Temperature dependence of resistivity $\rho(T)$ of semiconductors, insulators and metals is significantly based on the following factors</p>
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<p>(a) number of charge carriers can change with temperature $T$</p>
<p>(b) time interval between two successive collisions can depend on $T$</p>
<p>(c) length of material can be a function of $T$</p>
<p>(d) mass of carriers is a function of $T$</p>
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<p><strong>Ques.</strong> The measurement of an unknown resistance $R$ is to be carried out using Wheatstones bridge. Two students perform an experiment in two ways. The first students takes $R_2=10 \Omega$ and $R_1=5 \Omega$. The other student takes $R_2=1000 \Omega$ and $R_1=500 \Omega$. In the standard arm, both take $R_3=5 \Omega$.</p>
<p>Both find $R=\frac{R_2}{R_1}, R_3=10 \Omega$ within errors.</p>
<div class="fragment">
<p>(a) The errors of measurement of the two students are the same</p>
<p>(b) Errors of measurement do depend on the accuracy with which $R_2$ and $R_1$ can be measured</p>
<p>(c) If the student uses large values of $R_2$ and $R_1$ the currents through the arms will be feeble. This will make determination of null point accurately more difficult</p>
<p>(d) Wheatstone bridge is a very accurate instrument and has no errors of measurement</p>
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<p><strong>Ans.</strong></p>
<p>When an electron approaches a junction, in addition to the uniform electric field $E$ facing it normally. It keep the drift velocity fixed as drift velocity depend on $E$ by the relation drift velocity</p>
<p>$v_d=\frac{e E \tau}{m}$</p>
<p>This result into accumulation of charges on the surface of wires at the junction. These produce additional electric field. These fields change the direction of momentum.</p>
<p>Thus, the motion of a charge across junction is not momentum conserving.</p>
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<p><strong>Thinking Process</strong></p>
<p>The higher drift velocities of electrons make collisions more frequent which in turn decreases the time interval between two successive collision.</p>
<p><strong>Ans.</strong></p>
<p>Relaxation time is inversely proportional to the velocities of electrons and ions. The applied electric field produces the insignificant change in velocities of electrons at the order of $1 mm / s$, whereas the change in temperature $(T)$, affects velocities at the order of $10^{2} m / s$. This decreases the relaxation time considerably in metals and consequently resistivity of metal or conductor increases as .</p>
<p>$\rho=\frac{1}{\sigma}=\frac{m}{n e^{2} \tau}$</p>
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<p><strong>Ans.</strong></p>
<p>The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer.</p>
<p>It is easy and convenient method for observer.</p>
<p>The $R   _{\text{unknown }}$ can be calculated applying Kirchhoff’s rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.</p>
<p>Note The necessary and sufficient condition for balanced Wheatstone bridge is</p>
<p>$\frac{P}{Q}=\frac{R}{S}$</p>
<p>where $P$ and $Q$ are ratio arms and $R$ is known resistance and $S$ is unknown resistance.</p>
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<p><strong>Ans.</strong></p>
<p>The power consumption in transmission lines is given by $P=i^{2} R_C$, where $R_C$ is the resistance of transmission lines. The power is given by</p>
<p>$P=V I$</p>
<p>The given power can be transmitted in two ways namely (i) at low voltage and high current or (ii) high voltage and low current. In power transmission at low voltage and high current more power is wasted as $P \propto i^{2}$ whereas power transmission at high voltage and low current facilitates the power transmission with minimal power wastage.</p>
<p>The power wastage can be reduced by transmitting power at high voltage.</p>
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<p><strong>Ans.</strong></p>
<p>With the increase of $R$, the current in main circuit decreases which in turn, decreases the potential difference across $A B$ and hence potential gradient $(k)$ across $A B$ decreases.</p>
<p>Since, at neutral point, for given emf of cell, $I$ increases as potential gradient (k) across $A B$ has decreased because</p>
<p>$E=k I$</p>
<p>Thus, with the increase of $I$, the balance point neutral point will shift towards $B$.</p>
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<p><strong>Ques.</strong> While doing an experiment with potentiometer (figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one and $A$ of the wire, to the end $R$; (ii) the deflection increased, while the jockey was moved towards the end $D$.</p>
<img src="https://cdn.mathpix.com/snip/images/1eJUROx7EmJRufLvLZO7dWQyXfz3sObwWml_ny6PURE.original.fullsize.png" height="120px">
<p>(i) Which terminal positive or negative of the cell $E_1$, is connected at $X$ in case (i) and how is $E_1$ related to $E$ ?</p>
<p>(ii) Which terminal of the cell $E_1$ is connected at $X$ in case (ii))?</p>
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<p><strong>Ans.</strong></p>
<p>(i) The deflection in galvanometer is one sided and the deflection decreased, while moving from one end ‘$A$’ of the wire to the end ‘$B$’, thus imply that current in auxiliary circuit (lower circuit containing primary cell) decreases, while potential difference across $A$ and jockey increases.</p>
<p>This is possible only when positive terminal of the cell $E_1$, is connected at $X$ and $E_1>E$.</p>
<p>(ii) The deflection in galvanometer is one sided and the deflection increased, while moving from one end $A$ of the wire to the end $B$, this imply that current in auxiliary circuit (lower circuit containing primary cell) increases, while potential difference across $A$ and jockey increases.</p>
<p>This is possible only when negative terminal of the cell $E_1$, is connected at $X$.</p>
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<p><strong>Ans.</strong></p>
<p>When all resistances are connected in parallel, the resultant resistance $R_P$ is given by</p>
<p>$\frac{1}{R_P}=\frac{1}{R_1}+\ldots \ldots \ldots+\frac{1}{R_n}$</p>
<p>On multiplying both sides by $R   _{\min }$ we have</p>
<p>$\frac{R   _{\min }}{R_P}=\frac{R  _{\min }}{R_1}+\frac{R  _{\min }}{R_2}+\ldots . .+\frac{R  _{\min }}{R_n}$</p>
<p>Here, in RHS, there exist one term $\frac{R   _{\text{min }}}{R _{\text{min }}}=1$ and other terms are positive, so we have</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$\frac{R   _{\min }}{R_p}=\frac{R  _{\min }}{R_1}+\frac{R  _{\min }}{R_2}+\ldots .+\frac{R  _{\min }}{R_n}>1$</p>
<p>This shows that the resultant resistance $R_p<R   _{\text{min }}$.</p>
<p>Thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors. Now, in series combination, the equivalent resistant is given by</p>
<p>$R_s=R_1+\ldots \ldots+R_n$</p>
<p>Here, in RHS, there exist one term having resistance $R   _{\max }$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>In Fig. (b), $R   _{\text{min }}$ provides an equivalent route as in Fig. (a) for current. But in addition there are $(n-1)$ routes by the remaining $(n-1)$ resistors. Current in Fig. (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) $<R _{\text{min }}$. Second circuit evidently affords a greater resistance.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>In Fig. (d), $R   _{\max }$ provides an equivalent route as in Fig. (c) for current. Current in Fig. (d) < current in Fig. (c). Effective resistance in Fig. (d) $>R  _{\max }$. Second circuit evidently affords a greater resistance.</p>
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<p><strong>Thinking Process</strong></p>
<p>Here, after finding the electric current flow in the circuit by using Kirchhoff ’s law or Ohm’s law, the potential difference across first cell can be obtained.</p>
<p><strong>Ans.</strong></p>
<p>Applying Ohm’s law,</p>
<p>Effective resistance $=R+r_1+r_2$ and effective emf of two cells $=E+E=2 E$,</p>
<p>So the electric current is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking Process</strong></p>
<p>The electric current in two cases is obtained and then shown equal to each other</p>
<p><strong>Ans.</strong></p>
<p>Let the effective internal resistance of the battery is $R   _{\text{eff }}$, the effective external resistance $R$ and the effective voltage of the battery is $V _{\text{eff }}$.</p>
<p>Applying Ohm’s law,</p>
<p>Then current through $R$ is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> In an experiment with a potentiometer, $V_B = 10V$. $R$ is adjusted to be $50 \Omega$. A student wanting to measure voltage $E_1$ of a battery (approx. 8V) finds no null point possible. He then diminishes $R$ to $10 \Omega$ and is able to locate the null point on the last $(4^\text{th})$ segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.</p>
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<p><strong>Thinking Process</strong></p>
<p>The null point is obtained only when emf of primary cell is less than the potential difference across the wires of potentiometer.</p>
<p><strong>Ans.</strong></p>
<p>Let $R^{\prime}$ be the resistance of the potentiometer wire.</p>
<p>Effective resistance of potentiometer and variable resistor $(R=50 \Omega)$ is given by $=50 \Omega+R^{\prime}$</p>
<p>Effective voltage applied across potentiometer $=10 V$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$ \frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}}  <8 $</p>
<p>$ \Rightarrow  7.5 R^{\prime}  <80+8 R^{\prime} $</p>
<p>$ R^{\prime}  >160 $</p>
<p>$ 160  <R^{\prime}<200 $</p>
<p>Any $R^{\prime}$ between $160 \Omega$ and $200 \Omega$ will achieve.</p>
<p>Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across $400 cm$ of wire $>8 V$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> (a) Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (Ignore thermal motion) to the state of drift velocity?</p>
<p>(b) Electrons give up energy at the rate of $RI^{2}$ per second to the thermal energy. What time scale would number associate with energy in problem (a)? $n$ = number of electron/volume = $10^{29} / m^{3}$, length of circuit = 10 cm, cross-section = A = $(1 mm)^{2}$.</p>
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<p><strong>Thinking Process</strong></p>
<p>The current in a conductor and drift velocity of electrons are related as $i=n e A v_d$, where $v_d$ is drift speed of electrons and $n$ is number density of electrons.</p>
<p><strong>Ans.</strong></p>
<p>(a) By Ohm’s law, current $I$ is given by</p>
<p>$ I=6 V / 6 \Omega=1 A $</p>
<p>$ I=n e t A v_d \text{ or } v_d=\frac{i}{n e A} $</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$ =\frac{1}{1.6} \times 10^{-4} m / s $</p>
<p>Therefore, the energy absorbed in the form of KE is given by</p>
<p>$ KE  =\frac{1}{2} m_e v_d^{2} \times n A I \newline $</p>
<p>$ =\frac{1}{2} \times 9.1 \times 10^{31} \times \frac{1}{2.56} \times 10^{20} \times 10^{8} \times 10^{6} \times 10^{1} \newline $</p>
<p>$ =2 \times 10^{-17} J $</p>
<p>(b) Power loss is given by $P=I^{2} R=6 \times 1^{2}=6 W=6 J / s$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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