 # 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 :

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

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

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 :

b) Current density of wire

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

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

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 :

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

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

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

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

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 :

$\therefore$ 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

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.

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

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.

19   21.

##### Solution :

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

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

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 :

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

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

30   32.

##### Solution :

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