Electric Current in Conductors

Concept Of Physics

H C Verma

1   1.The amount of charge passed in time t through a cross-section of a wire is $Q(t) =At^2 + Bt + C.\\$ (a) Write the dimensional formulae for A, B and C.$\\$ (b) If the numerical values of A, B and C are 5, 3 and 1 respectively in SI units, find the value of the current at t = 5 s.

Solution :

$Q(t)=At^2+Bt+c\\ a)\ At^2=Q\\ \Rightarrow A=\frac{Q}{t^2}=\frac{A'T'}{T^{-2}}=A^1T^{-1}\\ b)\ Bt=Q\\ \Rightarrow B=\frac{Q}{T}=\frac{A'T'}{T}=A\\ c)\ C=[Q]\\ \Rightarrow C=A'T'\\ d) Current\ t=\frac{dQ}{dt}=\frac{d}{dt}(At^2+Bt+C)\\ =2At+B=2\times 5\times 5+3=53\ A.$

2   2. An electron gun emits $2.0 \times 10^{19}$ electrons per second. What electric current does this correspond to?

Solution :

$ No.\ of\ electrons\ per\ second\ = 2 \times 10^{16}\ electrons / sec.\\ Charge\ passing\ per\ second\ = 2 \times 10^{16}\times 1.6 \times 10^{–9} \frac{coulomb}{sec}\\ = 3.2 \times 10^{–9} Coulomb/sec\\ Current = 3.2 \times 10^{–3} A.$

3   3. The electric current existing in a discharge tube is $2.0\ \mu A$. How much charge is transferred across a cross-section of the tube in 5 minutes ?

Solution :

$i' = 2 \mu A,\ t = 5\ min = 5 \times 60\ sec.\\ q = i t = 2 \times 10^{–6} \times 5 \times 60\\ = 10 \times 60 \times 10^{–6}\ c = 6 \times 10^{–4}\ c$

4   4. The current through a wire depends on time as $i = i_0 + at,$ where $i_0 = 10\ $ A and a = 4 A/s. Find the charge crossed through a section of the wire in 10 seconds.

Solution :

$i=i_0+\alpha t,\ t=10\ sec, i_0=10\ A,\alpha=4\ A/sec.\\ q=\int_{0}^{t} idt =\int_{0}^{t} (i_0+\alpha t)dt=\int_{0}^{t} i_0 dt+\int_{0}^{t} \alpha tdt $

$=i_0 t+\alpha\frac{t^2}{2}=10\times 10+4\times \frac{10\times 10}{2}\\ =100+200=300\ C$

5   5. A current of 1.0 A exists in a copper wire of cross-section $1.0\ mm^2$. Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is 9000 $kg/m^3$.

Solution :

6   6. A wire of length 1 m and radius 0.1 mm has a resistance of 100 $\Omega $. Find the resistivity of the material.

Solution :

$l = 1\ m,\ r = 0.1\ mm\ = 0.1 \times 10^{–3}\ m\\ R = 100 \Omega ,\ f = ?\\ \Rightarrow R = \frac{fl}{a}\\ \Rightarrow f = \frac{Ra}{l}=\frac{100\times 3.14\times 0.1\times 0.1\times 10^{-6}}{1}\\ = 3.14 \times 10^{–6} = \pi \times 10^{–6} \Omega -m.$

7   7. A uniform wire of resistance 100 Q is melted and recast in a wire of length double that of the original. What would be the resistance of the wire ?

Solution :

$l'= 2\ l\\ volume\ of\ the\ wire\ remains\ constant.\\ A\ l = A'l'\\ \Rightarrow A\ l = A'\times 2\ l\\ \Rightarrow A' = \frac{A}{2}\\ f = Specific\ resistance\\ R = \frac{fl}{A}\ ; R' = \frac{fl'}{A'}\\ 100\Omega = \frac{f2l}{A/2} =\frac{4fl}{A} = 4R\\ \Rightarrow 4 \times 100 \Omega = 400 \Omega $

8   8. Consider a wire of length 4 m and cross-sectional area 1 mm carrying a current of 2 A. If each cubic metre of the material contains $10^{29}$ free electrons, find the average time taken by an electron to cross the length of the wire.

Solution :

$l = 4\ m,\ A = 1 mm^2 = 1 \times 10^{–6}\ m^2\\ I = 2 A,\ \frac{n}{V} = 10^{29},\ t = ?\\ i = n A V_d e\\ \Rightarrow e = 10^{29} \times 1 \times 10^{–6} \times V_d \times 1.6 \times 10^{–19}\\ \Rightarrow V_d=\frac{2}{10^{29}\times 10^{-6}\times 1.6\times 10^{-19}}\\ =\frac{1}{0.8\times 10^4}=\frac{1}{8000}\\ t=\frac{l}{V_d}=\frac{4}{1/8000}=4\times 8000$

9   9. What length of a copper wire of cross-sectional area 0.01 mm will be needed to prepare a resistance of 1 kQ ? Resistivity of copper - $1.7 \times 10^{-8}\ n-m.$

Solution :

$f_{cu} = 1.7 \times 10^{–8} \Omega -m\\ A = 0.01\ mm^2 = 0.01 \times 10^{–6} m^2\\ R = 1\ K\Omega = 10^3 \Omega\\ R=\frac{fl}{a}\\ \Rightarrow 10^3=\frac{1.7\times 10^{-8}\times l}{10^{-6}}\\ \Rightarrow l=\frac{10^3}{1.7}=0.58\times 10^3\ m=0.6\ km.$

10   11. A copper wire of radius 01 mm and resistance $1\ k\Omega $ is connected across a power supply of 20 V. $\\$ (a) How many electrons are transferred per second between the supply and the wire at one end ? $\\$ (b) Write down the current density in the wire.

Solution :

$r = 0.1\ mm = 10^{–4} m\\ R = 1 K\Omega = 10^3 \Omega,\ V = 20 V\\ a)\ No.\ of\ electrons\ transferred\\ i = \frac{V }{R}= \frac{20}{10^3} = 20 \times 10^{–3} = 2 \times 10^{–2} A\\ q=it=2\times 10^{-2} \times 1=2\times 10^{-2}\ C.\\ No.\ of\ electrons\ transferred=\frac{2\times 10^{-2}}{1.6\times10^{-19}}=\frac{2\times 10^{-17}}{1.6}=1.25\times 10^{17}$

$= \frac{i}{A}=\frac{2\times 10^{-2}}{\pi \times 10^{-8}} =\frac{2}{3.14}\times 10^6\\ = 0.6369 \times 10^6 = 6.37 \times 10^5 A/m^2.$

b) Current density of wire

11   12. Calculate the electric field in a copper wire of cross-sectional area 2'0 mm2 carrying a current of 1 A. The resistivity of copper = $1.7 \times 10^{-8} \Omega -m.$

Solution :

$A = 2 \times 10^{–6}\ m^2,\ I = 1\ A\\ f = 1.7 \times 10^{–8} \Omega -m\\ E =?\\ R = \frac{fl}{A}=\frac{1.7\times 10^{-8} \times l}{2\times 10^{-6}}\\ V = IR = \frac{1\times 1.7\times10^{-8}\times l}{2\times10^{-6}}\\ E = \frac{dV}{dL}=\frac{V}{I}=\frac{1.7\times 10^{-8}\times l}{2\times 10^{-6}l}=\frac{1.7}{2}\times 10^{-2}\ V / m\\ = 8.5\ mV/m.$

12   13. A wire has a length of 2.0 m and a resistance of 5.0 $\Omega$. Find the electric field existing inside the wire if it carries a current of 10 A.

Solution :

$I = 2\ m,\ R = 5 \Omega ,\ i = 10\ A,\ E = ?\\ V\ = iR\ = 10 \times 5 = 50\ V\\ E = \frac{V}{I}=\frac{50}{2} = 25\ V/m.$

13   14. The resistances of an iron wire and a copper wire at 20°C are 3.9 $\Omega $ and 4.1 $\Omega $ respectively. At what temperature will the resistances be equal ? Temperature coefficient of resistivity for iron is $5.0 \times 10^{-3}$ and for copper it is $4.0 \times 10^{-3}$ K Neglect any thermal expansion.

Solution :

$R'_{Fe} = R_{Fe} (1 + \alpha_{Fe}\Delta\theta), R'_{Cu} = R_{Cu} (1 + \alpha_{Cu}\Delta\theta)\\ R'_{Fe} = R'_{Cu}\\ \Rightarrow R_{Fe}(1 + \alpha_{Fe}\Delta\theta), = R_{Cu} (1 + \alpha_{Fe}\Delta\theta)$

$ \Rightarrow 3.9 + 3.9 \times 5 \times 10{–3} (20 – \theta) = 4.1 + 4.1 \times 4 \times 10^{–3} (20 – \theta)\\ \Rightarrow 4.1 \times 4 \times 10^{–3} (20 – \theta) – 3.9 \times 5 \times10^{–3} (20 – \theta) = 3.9 – 4.1\\$

$\Rightarrow 3.9 [ 1 + 5 \times 10^{–3} (20 – \theta)] = 4.1 [1 + 4 \times 10^{–3} (20 – \theta)]\\$

$\Rightarrow 16.4(20 – \theta) – 19.5(20 – \theta) = 0.2 \times 10^3\\ \Rightarrow (20 – \theta) (–3.1) = 0.2 \times 10^3\\ \Rightarrow \theta – 20 = 200\\ \Rightarrow \theta = 220°C.$

14   15. The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has a zero error (that is, it does not read zero when no potential difference is applied). Calculate the zero error if the readings for two different conditions are 1.75 A, 14.4 V and 2.75 A, 22.4 V.

Solution :

$Let\ the\ voltmeter\ reading\ when,\ the\ voltage\ is\ 0\ be\ X.\\ \frac{I_1R}{I_2R}=\frac{V_1}{V_2}\\ \Rightarrow \frac{1.75}{2.75}=\frac{14.4-V}{22.4-V}\Rightarrow \frac{0.35}{0.55}=\frac{14.4-V}{22.4-V}\\ \Rightarrow \frac{0.07}{0.11}=\frac{14.4-V}{22.4-V}\Rightarrow \frac{7}{11}=\frac{14.4-V}{22.4-V}\\ \Rightarrow 7(22.4-V)=11(14-4-V)\Rightarrow 156.8-7V=158.4-11V\\ \Rightarrow (7 – 11)V = 156.8 – 158.4 \Rightarrow –4V = –1.6\\ \Rightarrow V = 0.4\ V.$

15   16. Figure (32-E2) shows an arrangement to measure the emf E and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1'52 V when the switch S is open. When the switch is closed the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.

Solution :

a) When switch is open, no current passes through the ammeter. In the upper part of the circuit the Voltmenter has $\infty$ resistance. Thus current in it is 0.

$E – ir = 1.45\\ \Rightarrow 1.52 – ir = 1.45\\ \Rightarrow ir = 0.07\\ \Rightarrow 1\ r = 0.07 \Rightarrow r = 0.07 \Omega .$

$\therefore$ Voltmeter read the emf. (There is not Pot. Drop across the resistor).$\\$ b) When switch is closed current passes through the circuit and if its value of i. The voltmeter reads

16   17. The potential difference between the terminals of a battery of emf 60 V and internal resistance $1\ \Omega$ drops to $5.8\ V$ when connected across an external resistor. Find the resistance of the external resistor.

Solution :

$E = 6\ V,\ r = 1 \Omega,\ V = 5.8\ V,\ R = ? \\ I = \frac{E}{R+r} =\frac{6}{R+1},\ V = E – Ir\\ \Rightarrow 5.8 = 6 - \frac{6}{R+1} \times 1 \Rightarrow \frac{6}{R+1} = 0.2\\ \Rightarrow R+1 =30 \Rightarrow R = 29 \Omega.$

17   18. The potential difference between the terminals of a 6.0 V battery is 7.2 V when it is being charged by a current of 2.0 A. What is the internal resistance of the battery ?

Solution :

$V = E+ ir\\ \Rightarrow 7.2 = 6 + 2 \times r\\ \Rightarrow 1.2 = 2r \Rightarrow r = 0.6 \Omega.$

18   19. The internal resistance of an accumulator battery of emf 6 V is 10 $\Omega $ when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1 ft. The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9 V. Find the current through the battery $\\$(a) just after the connections are made and $\\$(b) after a long time when it is completely charged.

Solution :

19   21.

Solution :

$a)\ Total\ emf\ = n_1E\\ in\ 1\ row\\ Total\ emf\ in\ all\ news\ = n_1E\\ Total\ resistance\ in\ one\ row\ =\ n_1r\\ Total\ resistance\ in\ all\ rows\ =\frac{n_1r}{n_2}\\ Net\ resistance\ = \frac{n_1r}{n_2} + R\\ Current\ =\frac{n_1E}{n_1/n_2r+R}=\frac{n_1n_2E}{n_1r+n_2R}$

$b) I=\frac{n_1n_2E}{n_1r+n_2R}\\ for\ I=max,\\ n_1r+n_2R=min\\ \Rightarrow (\sqrt{n_1r}-\sqrt{n_2R})^2+2\sqrt{n_1n_2rR}=min\\ it\ is\ min\ when\\ \sqrt{n_1r}=\sqrt{n_2R}\\ \Rightarrow n_1r=n_2R\\ I\ is\ max\ when\ n_1r=n_2r.\\$

20   22.

Solution :

$E = 100\ V,\ R' = 100 k\Omega = 100000\\ R = 1 – 100\\ When\ no\ other\ resister\ is\ added\ or\ R = 0.\\ i = \frac{E}{R'} =\frac{100 }{100000} =0.001\ Amp\\ when\ R=1\\ i=\frac{100}{100000+1} =\frac{100}{100001} =0.0009A\\ when\ R=100\\ i=\frac{100}{100000+100}=\frac{100}{100100}= 0.000999 A .\\ Upto\ R\ = 100\ the\ current\ does\ not\ upto\ 2\ significant\ digits.\ Thus\ it\ proved.$

21   23.

Solution :

$A_1 = 2.4\ A\\ Since\ A_1\ and\ A_2\ are\ in\ parallel,\\ \Rightarrow 20 \times 2.4 = 30 \times X \Rightarrow X = \frac{20 \times 2.4}{30} = 1.6\ A.\\ Reading\ in\ Ammeter\ A_2\ is\ 1.6\ A.\\ A_3 = A_1 + A_2 = 2.4 + 1.6 = 4.0\ A.$

22   24.

Solution :

$i_{min} = \frac{5.5\times 3}{110} = 0.15\\ i_{max}=\frac{5.5\times 3}{20}=\frac{16.5}{20}=0.825$

23   25.

Solution :

$a)\ R_{eff}=\frac{180}{3}=60\ \Omega\\ i=60/60=1\ A\\ b)\ R_{eff}=\frac{180}{2}=90\ \Omega\\ i=60/90=0.67\ A\\ c)\ R_{eff}=180\ \Omega\\ i=60/80=0.33\ A$

24   26.

Solution :

$Max.\ R\ = (20 + 50 + 100) \Omega = 170\ \Omega\\ Min\ R\ = \frac{1}{(\frac{1}{20}+\frac{1}{50}+\frac{1}{100})} = \frac{100}{8} = 12.5 \Omega.$

25   27.

Solution :

$The\ various\ resistances\ of\ the\ bulbs\ = \frac{V^2}{P}\\ Resistances\ are\ \frac{(15)^2}{10},\frac{ (15)^2}{10},\frac{(15)^2}{15}=45,\ 22.5\ 15.\\ Since\ two\ resistances\ when\ used\ in\ parallel\ have\ resistances\\ less\ than\ both.\ The\ resistances\ are\ 45\ and\ 22.5.$

26   28.

Solution :

$i_1 \times 20 = i_2 \times 10\\ \Rightarrow \frac{i_1}{i_2}=\frac{10}{20}=\frac{1}{2}\\ i_1 =\ 4\ mA,\ i_2 =\ 8\ mA\\ Current\ in\ 20\ K\Omega\ resistor = 4\ mA\\ Current\ in\ 10 K\Omega\ resistor\ = 8\ mA\\ Current\ in\ 100 K\Omega\ resistor\ =\ 12\ mA\\ V = V_1 + V_2 + V_3\\ = 5\ K\Omega \times 12\ mA + 10 K\Omega \times 8\ mA + 100 K\Omega \times 12\ mA\\ = 60 + 80 + 1200 = 1340\ volts$

27   29.

Solution :

$R_1 = R,\ i_1 = 5\ A\\ R_2\ =\frac{10R}{10+R},\ i_2 =\ 6A\\ Since\ potential\ constant,\\ i_1R_1 = i_2R_2\\ \Rightarrow 5 \times R = \frac{6\times 10R}{10+R}\\ \Rightarrow (10 + R)5 = 60\\ \Rightarrow 5R = 10 \Rightarrow R = 2 \Omega.$

28   30.

Solution :

$Eq.\ Resistance\ = \frac{r}{3}.$

29   31.

Solution :

$a)\ R_{eff}=\frac{\frac{15\times 5}{6}\times \frac{15}{6}}{\frac{15\times 5}{6}+\frac{15}{6}}=\frac{\frac{15\times 5\times 15}{6\times 6}}{\frac{75+15}{6}}$

$=\frac{15\times 5\times 15}{6\times 90}=\frac{25}{12}=2.08\ \Omega\\ b)\ Across AC,\\ R_{eff}=\frac{\frac{15\times 4}{6}\times \frac{15\times 2}{6}}{\frac{15\times 4}{6}+\frac{15\times 2}{6}}=\frac{\frac{15\times 4\times 15\times 2}{6\times 6}}{\frac{60+30}{6}}\\ =\frac{15\times 4\times 15\times 2}{6\times 90}=\frac{10}{3}=3.33\ \Omega\\ c)\ Across AD,\\ R_{eff}=\frac{\frac{15\times 3}{6}\times \frac{15\times 3}{6}}{\frac{15\times 3}{6}+\frac{15\times 3}{6}}=\frac{\frac{15\times 3\times 15\times 3}{6\times 6}}{\frac{60+30}{6}}\\ =\frac{15\times 3\times 15\times 3}{6\times 90}=\frac{15}{4}=3.75\ \Omega$

30   32.

Solution :

a) when S is open$\\ Req\ = (10 + 20) \Omega = 30 \Omega.\\ i = When\ S\ is\ closed,\\ Req\ = 10 \Omega\\ i = (3/10) \Omega = 0.3 \Omega.$