Current Electricity
Chapter 3
CURRENT ELECTRICITY
MCQ I
3.1 Consider a current carrying wire (current $I$ ) in the shape of a circle. Note that as the current 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 is
(a) source of emf.
(b) electric field produced by charges accumulated on the surface of wire.
(c) the charges just behind a given segment of wire which push them just the right way by repulsion.
(d) the charges ahead.
Show Answer
Answer: (b)3.2 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 Fig 3.1.
(a) The equivalent emf $\varepsilon _{\text{eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e. $\varepsilon_1<\varepsilon _{eq}<\varepsilon_2$.
Fig 3.1
(b) The equivalent emf $\varepsilon _{\text{eq }}$ is smaller than $\varepsilon_1$.
(c) The $\varepsilon _{\text{eq }}$ is given by $\varepsilon _{\text{eq }}=\varepsilon_1+\varepsilon_2$ always.
(d) $\varepsilon _{\text{eq }}$ is independent of internal resistances $r_1$ and $r_2$.
Show Answer
Answer: (a)3.3 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 $l_1=2.9 cm$. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure $l_1$ more accurately.
(b) He should change $S$ to $1000 \Omega$ and repeat the experiment.
(c) He should change $S$ to $3 \Omega$ and repeat the experiment.
(d) He should give up hope of a more accurate measurement with a meter bridge.
Show Answer
Answer: (c)3.4 Two cells of emf’s approximately $5 V$ and $10 V$ are to be accurately compared using a potentiometer of length $400 cm$.
(a) The battery that runs the potentiometer should have voltage of $8 V$.
(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 $10 V$.
(c) The first portion of $50 cm$ of wire itself should have a potential drop of $10 V$.
(d) Potentiometer is usually used for comparing resistances and not voltages.
Show Answer
Answer: (b)3.5 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
(a) maximum when the battery is connected across $1 cm \times \frac{1}{2} cm$ faces.
(b) maximum when the battery is connected across $10 cm \times 1 cm$ faces.
(c) maximum when the battery is connected across $10 cm \times \frac{1}{2}$ $cm$ faces.
(d) same irrespective of the three faces.
Show Answer
Answer: (a)3.6 Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone.
(b) Thermal velocity alone.
(c) Both drift velocity and thermal velocity.
(d) Neither drift nor thermal velocity.
MCQ II
Show Answer
Answer: (a)3.7 Kirchhoff’s junction rule is a reflection of
(a) conservation of current density vector.
(b) conservation of charge.
(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) the fact that there is no accumulation of charges at a junction.
Show Answer
Answer: (b), (d)3.8 Consider a simple circuit shown in Fig 3.2. stands for a variable resistance $R^{\prime}$. $R^{\prime}$ can vary from $R_0$ to infinity. $r$ is internal resistance of the battery $(r«R«R_0)$.
(a) Potential drop across $AB$ is nearly constant as $R^{\prime}$ is varied.
(b) Current through $R^{\prime}$ is nearly a constant as $R^{\prime}$ is varied.
(c) Current $I$ depends sensitively on $R^{\prime}$.
(d) $I \geq \frac{V}{r+R}$ always.
Fig 3.2
Show Answer
Answer: (a), (d)3.9 Temperature dependence of resistivity $\rho(T)$ of semiconductors, insulators and metals is significantly based on the following factors:
(a) number of charge carriers can change with temperature $T$.
(b) time interval between two successive collisions can depend on $T$.
(c) length of material can be a function of $T$.
(d) mass of carriers is a function of $T$.
Show Answer
Answer: (a), (b)3.10 The measurement of an unknown resistance $R$ is to be carried out using Wheatstones bridge (Fig. 3.25). 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 $R_3=5 \Omega$.
Both find $R=\frac{R_2}{R_1} R_3=10 \Omega$ within errors.
(a) The errors of measurement of the two students are the same.
(b) Errors of measurement do depend on the accuracy with which $R_2$ and $R_1$ can be measured.
(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.
(d) Wheatstone bridge is a very accurate instrument and has no errors of measurement.
Show Answer
Answer: (b), (c)3.11 In a meter bridge the point $D$ is a neutral point (Fig 3.3).
(a) The meter bridge can have no other neutral point for this set of resistances.
(b) When the jockey contacts a point on meter wire left of $D$, current flows to $B$ from the wire.
(c) When the jockey contacts a point on the meter wire to the right of $D$, current flows from $B$ to the wire through galvanometer.
(d) When $R$ is increased, the neutral point shifts to left.
Fig 3.3
VSA
Show Answer
Answer: (a), (c)3.12 Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?
Show Answer
Answer: When an electron approaches a junction, in addition to the uniform $\mathbf{E}$ that it normally faces (which keep the drift velocity $\boldsymbol{{}v}_d$ fixed), there are accumulation of charges on the surface of wires at the junction. These produce electric field. These fields alter direction of momentum.3.13 The relaxation time $\tau$ is nearly independent of applied $E$ field whereas it changes significantly with temperature $T$. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of $\rho$ with temperature. Elaborate why?
Show Answer
Answer: Relaxation time is bound to depend on velocities of electrons and ions. Applied electric field affects the velocities of electrons by speeds at the order of $1 mm / s$, an insignificant effect. Change in $T$, on the other hand, affects velocities at the order of $10^{2} m / s$. This can affect $\tau$ significantly.
[ $\rho=\rho(E, T)$ in which $E$ dependence is ignorable for ordinary applied voltages.]
3.14 What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R _{\text{unknown }}$ by any other method?
Show Answer
Answer: The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point and there is no need to determine current in resistances and galvanometer and the internal resistance of a galvanometer. $R _{\text{unknown }}$ 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.3.15 What is the advantage of using thick metallic strips to join wires in a potentiometer?
Show Answer
Answer: The metal strips have low resistance and need not be counted in the potentiometer length $l_1$ of the null point. One measures only their lengths along the straight segments (of lengths 1 meter each). This is easily done with the help of centimeter rulings or meter ruler and leads to accurate measurements.3.16 For wiring in the home, one uses $Cu$ wires or $Al$ wires. What considerations are involved in this?
Show Answer
Answer: Two considerations are required: (i) cost of metal, and (ii) good conductivity of metal. Cost factor inhibits silver. $Cu$ and $Al$ are the next best conductors.3.17 Why are alloys used for making standard resistance coils?
Show Answer
Answer: Alloys have low value of temperature co-efficient (less temperature sensitivity) of resistance and high resistivity.3.18 Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$. If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.
Show Answer
Answer: Power wasted $P_C=I^{2} R_C$
where $R_C$ is the resistance of the connecting wires.
$P_C=\frac{P^{2}}{V^{2}} R_C$
In order to reduce $P_C$, power should be transmitted at high voltage.
3.19 $AB$ is a potentiometer wire (Fig 3.4). If the value of $R$ is increased, in which direction will the balance point $J$ shift?
Fig 3.4
Show Answer
Answer: If $R$ is increased, the current through the wire will decrease and hence the potential gradient will also decrease, which will result in increase in balance length. So J will shift towards B.3.20 While doing an experiment with potentiometer (Fig 3.5) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end $A$ of the wire to the end $B$; (ii) the deflection increased. while the jockey was moved towards the end $B$.
Fig 3.5
(i) Which terminal +or -ve of the cell $E_1$, is connected at $X$ in case (i) and how is $E_1$ related to $E$ ?
(ii) Which terminal of the cell $E_1$ is connected at $X$ in case (ii)?
Show Answer
Answer: (i) Positive terminal of $E_1$ is connected at $X$ and $E_1>E$.
(ii) Negative terminal of $E_1$ is connected at $X$.
3.21 A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$. Plot a graph showing the variation of P.D. across R, verses $R$.
SA
Show Answer
Answer:
3.22 First a set of $n$ equal resistors of $R$ each are connected in series to a battery of emf $E$ and internal resistance $R$. A current $I$ is observed to flow. Then the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ’ $n$ ‘?
Show Answer
Answer: $ I=\frac{E}{R+n R} ; \frac{E}{R+\frac{R}{n}}=10 I$
$\frac{1+n}{1+\frac{1}{n}}=10=\frac{1+n}{n+1} n=n$
$\therefore n=10$.
3.23 Let there be $n$ resistors $R_1 \ldots \ldots \ldots \ldots R_n$ with $R _{\max }=\max (R_1 \ldots \ldots \ldots R_n)$ and $R _{\min }=\min {R_1 \ldots \ldots R_n}$. Show that when they are connected in parallel, the resultant resistance $R_p<R _{\min }$ and when they are connected in series, the resultant resistance $R_S>R _{\max }$. Interpret the result physically.
Show Answer
Answer: $ \frac{1}{R_p}=\frac{1}{R_1}+\ldots \ldots+\frac{1}{R_n}, \quad \frac{R _{\min }}{R_P}=\frac{R _{\min }}{R_1}+\frac{R _{\min }}{R_2}+\ldots \ldots+\frac{R _{\min }}{R_n}>1$
and $R_S=R_1+\ldots \ldots+R_n \geq R _{\max }$.
In Fig. (b), $R _{\min }$ 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) $>$ current in Fig. (a). Effective Resistance in Fig. (b) $<R _{\min }$. Second circuit evidently affords a greater resistance. You can use Fig. (c) and (d) and prove $R_s>R _{\max }$.
(a)
(b)
(c)
(d)
3.24 The circuit in Fig 3.6 shows two cells connected in opposition to each other. Cell $E_1$ is of emf $6 V$ and internal resistance $2 \Omega$; the cell $E_2$ is of emf $4 V$ and internal resistance $8 \Omega$. Find the potential difference between the points $A$ and $B$.
Fig 3.6
Show Answer
Answer: $ \quad I=\frac{6-4}{2+8}=0.2 A$
P.D. across $E_1=6-0.2 \times 2=5.6 V$
P.D. across $E_2=V _{AB}=4+0.2 \times 8=5.6 V$
Point $B$ is at a higher potential than $A$
3.25 Two cells of same emf $E$ but internal resistance $r_1$ and $r_2$ are connected in series to an external resistor $R$ (Fig 3.7). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero.
Fig. 3.7
Show Answer
Answer: $\quad I=\frac{E+E}{R+r_1+r_i}$
$V_1=E-I r_1=E-\frac{2 E}{r_1+r_2+R} r_1=0$
or $E=\frac{2 E r_1}{r_1+r_2+R}$
$1=\frac{2 r_1}{r_1+r_2+R}$
$r_1+r_2+R=2 r_1$
$R=r_1-r_2$
3.26 Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter $1 mm$. Conductor B is a hollow tube of outer diameter $2 mm$ and inner diameter $1 mm$. Find the ratio of resistance $R_A$ to $R_B$.
Show Answer
Answer: $R_A=\frac{\rho l}{\pi(10^{-3} \times 0.5)^{2}}$
$R_B=\frac{\rho l}{\pi[(10^{-3})^{2}-(0.5 \times 10^{-3})^{2}]}$
$\frac{R_A}{R_B}=\frac{(10^{-3})^{2}-(0.5 \times 10^{-3})^{2}}{(.5 \times 10^{-3})^{2}}=3: 1$
3.27 Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3 .7 in the NCERT Text Book for Class XII.
Show Answer
Answer: We can think of reducing entire network to a simple one for any branch $R$ as shown in Fig.
Then current through $R$ is $I=\frac{V _{\text{eff }}}{R _{\text{eff }}+R}$
Dimensionally $V _{\text{eff }}=V _{\text{eff }}(V_1, V_2, \ldots \ldots V_n)$ has a dimension of voltage and $R _{\text{eff }}=R _{e f f}(R_1, R_2, \ldots \ldots . R_m)$ has a dimension of resistance.
Therefore if all are increased $n$-fold
$V _{e f f}^{\text{new }}=n V _{\text{eff }}, R _{e f f}^{\text{new }}=n R _{\text{eff }}$
and $R^{\text{new }}=n R$.
Current thus remains the same.
3.28 Two cells of voltage $10 V$ and $2 V$ and internal resistances $10 \Omega$ and $5 \Omega$ respectively, are connected in parallel with the positive end of $10 V$ battery connected to negative pole of $2 V$ battery (Fig 3.8). Find the effective voltage and effective resistance of the combination.
Fig 3.8
Show Answer
Answer: Applying Kirchhoffs junction rule:
$I_1=I+I_2$
Kirchhoff’s loop rule gives:
$10=I R+10 I_1 \ldots(i)$
$2=5 I_2-R I=5(I_1-I)-R I$
$4=10 I_1-10 I-2 R I \ldots$. (ii)
(i) - (ii) $\Rightarrow 6=3 R I+10 I$ or, $2=I(R+\frac{10}{3})$
$2=(R+R _{eff}) I$ Comparing with $V _{\text{eff }}=(R+R _{e f f}) I$
and $V _{\text{eff }}=2 V$
$R _{eff}=\frac{10}{3} \Omega$.
3.29 A room has $AC$ run for 5 hours a day at a voltage of $220 V$. The wiring of the room consists of $Cu$ of $1 mm$ radius and a length of $10 m$. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
$[\rho _{cu}=1.7 \times 10 _{\Omega m}^{-8}, \rho _{Al}=2.7 \times 10^{-8} \Omega m]$
Show Answer
Answer: Power consumption $=2$ units $/$ hour $=2 KW=2000 J / s$
$I=\frac{P}{V}=\frac{2000}{220} \simeq 9 A$
Power loss in wire $=R I^{2} J / s$
$ \begin{aligned} & =\rho \frac{l}{A} I^{2}=1.7 \times 10^{-8} \times \frac{10}{\pi \times 10^{-6}} \times 81 J / s \\ & \simeq 4 J / s \\ & =0.2 % \end{aligned} $
Power loss in Al wire $=4 \frac{\rho _{A l}}{\rho _{C u}}=1.6 \times 4=6.4 J / s=0.32 %$
3.30 In an experiment with a potentiometer, $V_B=10 V$. $R$ is adjusted to be $50 \Omega$ (Fig. 3.9). 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.
Fig 3.9
Show Answer
Answer: Let $R^{\prime}$ be the resistance of the potentiometer wire.
$\frac{10 \times R^{\prime}}{50+R^{\prime}}<8 \Rightarrow 10 R^{\prime}<400+8 R^{\prime}$
$2 R^{\prime}<400$ or $R^{\prime}<200 \Omega$.
$\frac{10 \times R^{\prime}}{10+R^{\prime}}>8 \Rightarrow 2 R^{\prime}>80 \Rightarrow R^{\prime}>40$
$\frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}}<8 \Rightarrow 7.5 R^{\prime}<80+8 R^{\prime}$
$R^{\prime}>160 \Rightarrow 160<R^{\prime}<200$.
Any R’ between $160 \Omega$ and $200 \Omega$ will achieve.
Potential drop across $400 cm$ of wire $>8 V$.
Potential drop across $300 cm$ of wire $<8 V$.
$ \phi \times 400>8 V(\phi \to \text{ potential gradient }) $
$ \begin{aligned} & \phi \times 300<8 V \\ & \phi>2 V / m \end{aligned} $
$ <2 \frac{2}{3} V / m $
3.31 (a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?
(b) Electrons give up energy at the rate of $RI^{2}$ per second to the thermal energy. What time scale would one associate with energy in problem (a)? $n=$ no of electron $/$ volume $=10^{29} / m^{3}$, length of circuit $=10 cm$, cross-section $=A=(1 mm)^{2}$
Fig 3.10
Show Answer
Answer: (a) $I=\frac{6}{6}=1 A=n e v_d A$
$ \begin{aligned} v_d= & \frac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}=\frac{1}{1.6} \times 10^{-4} m / s \\ K . E & =\frac{1}{2} m_e v_d^{2} \times n A l \\ & =\frac{1}{2} \times 9.1 \times 10^{-31} \times \frac{1}{2.56} \times 10^{-8} \times 10^{29} \times 10^{-6} \times 10^{-1} \simeq 2 \times 10^{-17} J \end{aligned} $
(b) Ohmic loss $=R I^{2}=6 \times 1^{2}=6 J / s$
All of KE of electrons would be lost in $\frac{2 \times 10^{-17}}{6} s \simeq 10^{-17} s$
