Current Electricity
A wire of resistance R1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :
(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.
(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.
(C) The drift velocity does not depend on the applied potential difference to the conductor.
(D) The drift velocity of electron is inversely proportional to the length of the conductor.
(E) The drift velocity increases with the increase in the temperature of conductor.
Choose the correct answer from the options given below :
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $r_{1}$ and $r_{2}$ $\left(r_{1}>r_{2}\right)$. If the potential difference across the source of internal resistance $r_{1}$ is zero, then the value of R will be :
A battery of $6 \mathrm{~V}$ is connected to the circuit as shown below. The current I drawn from the battery is :
The current I in the given circuit will be :
A current of 15 mA flows in the circuit as shown in figure. The value of potential difference between the points A and B will be:
Which of the following physical quantities have the same dimensions?
An electric cable of copper has just one wire of radius 9 mm. Its resistance is 14 $\Omega$. If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius 3 mm connected in parallel, then the new resistance of the combination will be :
The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2$\Omega$. The value of internal resistance of each cell is
Resistance of the wire is measured as 2 $\Omega$ and 3 $\Omega$ at 10$^\circ$C and 30$^\circ$C respectively. Temperature co-efficient of resistance of the material of the wire is :
A 72 $\Omega$ galvanometer is shunted by a resistance of 8 $\Omega$. The percentage of the total current which passes through the galvanometer is :
The equivalent resistance between points A and B in the given network is :
An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :
Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, is :
If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:
A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for ${1 \over 3}$ deflection in the galvanometer. Which of the below is true for measuring value of G?
What will be the most suitable combination of three resistors A = 2$\Omega$, B = 4$\Omega$, C = 6$\Omega$ so that $\left( {{{22} \over 3}} \right)$$\Omega$ is equivalent resistance of combination?
Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 $\Omega$. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :
The current I flowing through the given circuit will be __________A.
Explanation:
All $9 ~\Omega$ resistances are in parallel
${R_{eq}} = 3\,\Omega $
$I = {6 \over 3}A = 2A$
An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.
Explanation:
${P_{100}} = {{{V^2}} \over {{R_{100}}}} \Rightarrow {R_{100}} = {{{V^2}} \over {{P_{100}}}}$
${P_{60}} = {{{V^2}} \over {{R_{60}}}} \Rightarrow {R_{60}} = {{{V^2}} \over {{P_{60}}}}$
${P_{net}} = {{{V^2}} \over {{R_{60}} + {R_{100}}}} = {{{P_{60}}{P_{100}}} \over {{P_{60}} + {P_{100}}}} = {{60 \times 100} \over {160}} = 37.5$
This power developed is proportional to resistance.
So, $P{'_{60}} = {P_{net}} \times {{60} \over {160}} = 37.5 \times {{60} \over {160}} \simeq 14\,W$
As shown in the figure, a potentiometer wire of resistance $20 \,\Omega$ and length $300 \mathrm{~cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4 \mathrm{~V}$. For a resistance '$R$' of resistance box introduced into the circuit, the null point for a cell of $20 \,\mathrm{mV}$ is found to be $60 \mathrm{~cm}$. The value of '$R$' is ___________ $\Omega .$
Explanation:
$l = 3m$, ${R_w} = 20\,\Omega $
${\varepsilon _0} = 4V$
${{4 \times 20} \over {20 + R}} \times {{60} \over {300}} = 20 \times {10^{ - 3}}$
${4 \over {20 + R}} = 5 \times {10^{ - 3}}$
$20 + R = 800$
$R = 780\,\Omega $
In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is $40 \mathrm{~cm}$. The balancing length, if the radius of the wire $\mathrm{AB}$ is doubled, will be ______________ $\mathrm{cm}$.
Explanation:
Even if the radius of wire is doubled, the balancing point would not change as ${x \over {l - x}} = {{{R_1}} \over {{R_2}}}$, which is not including a term of area.
In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance $30 \mathrm{~cm}$ from the left side as shown at point D. If R is $5.6$ $\mathrm{k} \Omega$, then the value of unknown resistance 'S' will be __________ $\Omega$.
Explanation:
${R \over S} = {{70} \over {30}}$
$S = {3 \over 7} \times 5.6 \times {10^3} = 2.4 \times {10^3}\,\Omega $
$ = 2400\,\Omega $
A $1 \mathrm{~m}$ long copper wire carries a current of $1 \mathrm{~A}$. If the cross section of the wire is $2.0 \mathrm{~mm}^{2}$ and the resistivity of copper is $1.7 \times 10^{-8}\, \Omega \mathrm{m}$, the force experienced by moving electron in the wire is ____________ $\times 10^{-23} \mathrm{~N}$.
(charge on electorn $=1.6 \times 10^{-19} \,\mathrm{C}$)
Explanation:
$I = ne{v_d}A$
$J = {E \over \rho }$
$F = eE = {{1.7 \times 1.6 \times {{10}^{ - 19}} \times {{10}^{ - 8}}} \over {2 \times {{10}^{ - 6}}}}$
$ = 136 \times {10^{ - 23}}$ N
A potentiometer wire of length $300 \mathrm{~cm}$ is connected in series with a resistance 780 $\Omega$ and a standard cell of emf $4 \mathrm{V}$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\, \mathrm{mV}$ is found to be $60 \mathrm{~cm}$. The resistance of the potentiometer wire is ____________ $\Omega$.
Explanation:
$l = 300$ cm
$\varepsilon = Kx$
$20 \times {10^{ - 3}} = \left( {{{4 \times R} \over {780 + R}} \times {1 \over {300}}} \right)60$
$R = 20$
Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40 \mathrm{~cm}$. Now an unknown resistance $x$ is connected in series with $\mathrm{P}$ and new balancing length is found to be $80 \mathrm{~cm}$ measured from the same end. Then the value of $x$ will be ____________ $\Omega$.
Explanation:
${P \over {40}} = {Q \over {60}}$ ...... (1)
${{P + x} \over {80}} = {Q \over {20}}$ ..... (2)
${P \over {P + x}} \times {{80} \over {40}} = {{20} \over {60}}$
${4 \over {4 + x}} \times 2 = {1 \over 3}$
$24 = 4 + x$
$x = 20$
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.
Explanation:
$E \propto I$
${{1.2} \over {1.8}} = {{36} \over {I'}}$
$I' = {3 \over 2} \times 36 = 54$ cm
$\Delta I = I' - I = 54 - 36 = 18$ cm
In the given figure, the value of Vo will be _____________ V.
Explanation:
Using Kirchhoff's junction rule.
${{2 - {V_0}} \over 1} + {{4 - {V_0}} \over 1} + {{6 - {V_0}} \over 1} = 0$
$12 - 3{V_0} = 0$
${V_0} = 4\,V$
Eight copper wire of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2 l$ have the same resistance $(R)$ then its diameter will be ____________ d.
Explanation:
$RAB = R$
$R = {1 \over 8}$ (Resistance of one wire)
$ = {1 \over 8}\rho {l \over {\pi {{{d^2}} \over 4}}} = {{\rho l} \over {2\pi {d^2}}}$
Resistance of copper wire of length $2l$ and diameter $x = R$
$\rho {{2l} \over {\pi {{{x^2}} \over 4}}} = R$
${{8\rho l} \over {\pi {x^2}}} = {{\rho l} \over {2\pi {d^2}}}$
$16{d^2} = {x^2}$
$x = 4d$
The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K1' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K1' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ $\Omega$.
Explanation:
Shortcut Method :
Internal Resistance of Unknown Battery
$r=\left(\frac{\ell_1-\ell_2}{\ell_2}\right) \mathrm{R}$
Where l1 means balanced length when key K1 is open
Where l2 means balanced length when key K1 is closed
Here l1 = 120 cm and l2 = 80 cm and R = 4 $\Omega $
$ \therefore $ $r = \left( {{{120 - 80} \over {80}}} \right) \times 4$ = 2 $\Omega $
Normal Method :
Original Potentiometer is $\to$
Here, AB is called potentiometer wire. Let length of AB = l and resistance = R
$\therefore$ (1) Current in the circuit (i) = ${{4} \over {R + {R_h}}}$
(2) Potential difference between A and B is
= VA $-$ VB = $\Delta$VAB = i $\times$ R
(3) ${{\Delta {V_{AB}}} \over {AB}} = {{Potential\,difference} \over {Length}}$ = Potential Gradient
$ \Rightarrow {{\Delta {V_{AB}}} \over l} = {{iR} \over l} = {4 \over {R + {R_h}}} \times {R \over l}$ Volt/m
Now, a circuit added to potentiometer wire.
When K1 is open then circuit look like the following.
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 120 = {{4R} \over {(R + {R_h})l}} \times 120$ ...... (1)
(2) From external circuit :
VA $-$ 1.5 $-$ 0 $\times$ r = VC
$\Rightarrow$ VA $-$ VC = 1.5V = $\Delta$VAC ...... (2)
Comparing equation (1) and (2), we get
${{4R} \over {(R + {R_h})l}} \times 120 = 1.5$ ..... (3)
When the key K1 is closed then circuit look like this $\to$
When null point is found at C then no current will flow through the circuit added to the potentiometer wire but i1 current will flow through circuit created using 1.5V with internal resistance r and 4$\Omega$ resistance.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ ....... (4)
(2) From external circuit :
${i_1} = {{1.5} \over {4 + r}}$
${V_A} - 1.5 + {i_1}r = {V_C}$
$ \Rightarrow {V_A} - {V_C} = 1.5 - {i_1}r = 1.5 - {{1.5r} \over {4 + r}}$ ..... (5)
Or we can use this also,
${V_A} - {i_1} \times 4 = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {i_1} \times 4 = {{1.5} \over {4 + r}} \times 4$
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 \times {2 \over 3} = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 - 1.5 \times {2 \over 3}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 \times {1 \over 3}$
$ \Rightarrow 3r = 4 + r$
$ \Rightarrow 2r = 4$
$ \Rightarrow r = 2\,\Omega $
Other Method :
Now this can be converted to single battery and single resistance using combination of battery rule,
We know,
${E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}$
and ${r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}$
$\therefore$ Here, ${E_{eq}} = {{1.5 \times 4 + 0 \times r} \over {4 + r}} = {6 \over {4 + r}}$
and ${r_{eq}} = {{4r} \over {4 + r}}$
$\therefore$ Circuit becomes
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ .... (4)
(2) From external circuit :
${V_A} - {6 \over {4 + r}} - 0 \times {{4r} \over {4 + r}} = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {6 \over {4 + r}}$ ...... (5)
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} \times 80 = {6 \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} \times 80 = {6 \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 \times {2 \over 3} = {6 \over {4 + r}}$
$ \Rightarrow 4 + r = 6$
$ \Rightarrow r = 2\,\Omega $
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $\Omega$ is used to measure the potential difference across 500 $\Omega$ resistor, the reading of the voltmeter will be ___________ V.
Explanation:
New ${R_{eff}} = {{2000 \times 500} \over {2500}} + 600\,\Omega = 1000\,\Omega $
$\Rightarrow$ Reading of voltmeter $ = {{400} \over {1000}} \times 20 = 8$ volts
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x $\times$ 10$-$3 $\Omega$ cm. The value of x is ____________. [Take $\pi$ = 3.14]
Explanation:
Resistance $ = \tan 45^\circ = 1\,\Omega $
$ \Rightarrow 1 = {{pI} \over A}$
$ \Rightarrow p = {{\pi {{(1.2\,cm)}^2}} \over {31.4\,cm}} = 1.44 \times {10^{ - 1}}$ $\Omega$ cm
$ \Rightarrow x = 144$
For the network shown below, the value of VB $-$ VA is ____________ V.
Explanation:
${V_B} - {V_A} = i \times 2$
$ = {{15} \over {1 + 2}} \times 2$
$ \Rightarrow {V_B} - {V_A} = 10$ volts
All resistances in figure are 1 $\Omega$ each. The value of current 'I' is ${a \over 5}$ A. The value of a is _________.
Explanation:
Let the current is i
Using Kirchhoff's law
$iR + {i \over 2}R + {i \over 4}R + {i \over 8}R = 3$
$i = {{3 \times 8} \over {15}} = {8 \over 5}A$
So $a = 8$
A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 $\Omega$. The galvanometer (G) shows null deflection when tapping key is at 43 cm mark from end A. If the end correction for end A is 2 cm, then the determined value of R will be ____________ $\Omega$.
Explanation:
${{43 + 2} \over {15}} = {{57} \over R}$
$R = {{57 \times 15} \over {45}} = 19\,\Omega $
Current measured by the ammeter (A) in the reported circuit when no current flows through 10 $\Omega$ resistance, will be ________________ A.
Explanation:
For ${I_{10}} = 0$
${R \over 3} = {4 \over 6}$
$ \Rightarrow R = 2\,\Omega $
$ \Rightarrow {I_A} = {{36 \times (6 + 9)} \over {6 \times 9}}$
$ = {{36 \times 15} \over {6 \times 9}} = 10$ A
The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 $\times$ 106 A/m2. The current through the outer portion of the wire between radial distances ${r \over 2}$ and r is x$\pi$ A; where x is __________.
Explanation:
$i = A \times j$
$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$
$ = {{3\pi {R^2}} \over 4} \times j$
$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 1.0 \times {10^6}$
$ = 12\,\pi $
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be $\sqrt {{x \over 2}} $. The value of x is __________.
Explanation:
${R_{net}} = {{ma} \over 3} + {a \over {2m}}$
$ = a\left[ {{m \over 3} + {1 \over {2m}} - {2 \over {\sqrt 6 }} + {2 \over {\sqrt 6 }}} \right]$
$ = a\left[ {{{\left( {\sqrt {{m \over 3}} - {1 \over {\sqrt {2m} }}} \right)}^2} + \sqrt {{2 \over 3}} } \right]$
This will be minimum when
$\sqrt {{m \over 3}} = {1 \over {\sqrt {2m} }}$
or $m = \sqrt {{3 \over 2}} $
so $x = 3$
A cell, shunted by a 8 $\Omega$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 $\Omega$ resistance. The value of internal resistance of the cell will be ____________ $\Omega$.
Explanation:
${{{\varepsilon _1}8} \over {{r_1} + 8}} =3c$
${{{\varepsilon _1}4} \over {{r_1} + 4}} =2c$
$ \Rightarrow {{2({r_1} + 4)} \over {{r_1} + 8}} = {3 \over 2}$
$ \Rightarrow {r_1} = 8\,\Omega $
The current density in a cylindrical wire of radius 4 mm is 4 $\times$ 106 Am$-$2. The current through the outer portion of the wire between radial distances ${R \over 2}$ and R is ____________ $\pi$ A.
Explanation:
$i = A \times j$
$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$
$ = {{3\pi {R^2}} \over 4} \times j$
$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 4 \times {10^6}$
$ = 48\pi $
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.
Explanation:
Volume is constant so on length doubled
Area is halved so
$R = \rho {l \over A}$ and $R' = \rho {{2l} \over {{A \over 2}}} = 4\rho {l \over A} = 4R$
So percentage increase will be
$R\% = {{4R - R} \over R} \times 100 = 300\% $
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.
Explanation:
$300 = {I^2}R \times 15$
$ \Rightarrow R = 5\,\Omega $
Now $I_2^2R{t_2}$
$ = 9 \times 5 \times 10$
$ = 450\,J$
The total current supplied to the circuit as shown in figure by the 5 V battery is ____________ A.
Explanation:
The equivalent circuit is
$ \Rightarrow I = {5 \over {2.5}} = 2\,A$
A potentiometer wire of length 10 m and resistance 20 $\Omega$ is connected in series with a 25 V battery and an external resistance 30 $\Omega$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is ${x \over {10}}$. The value of x is __________.
Explanation:
$\therefore$ $E = I \times \left( {{{20} \over 4}} \right) = {{25} \over {(30 + 20)}} \times \left( {{{20} \over 4}} \right)$
$ = {1 \over 2} \times 5 = 2.5$ volts
$ = {{25} \over {10}}$ volts
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.
Explanation:
At balancing point, we know that emf is proportional to the balancing length. i.e.,
emf $\propto$ balancing length
Now, let the emf's be 3$\varepsilon $ and 2$\varepsilon $.
$\Rightarrow$ 3$\varepsilon $ = k(75) ..... (1)
and 2$\varepsilon $ = k(l) ....... (2)
$\Rightarrow$ l = 50 cm
$\Rightarrow$ Difference is (75 $-$ 50) cm = 25 cm.
