# Electric current resistance and resistivity Problems

Question 1:
A cylindrical tube of length L has inner radius a and outer radius b as shown below in the figure

Given that ρ is the resistivity of the material. Find the resistance of the tube between its ends.

We know that current density at any point inside the conductor is
$J=\sigma E$
=> $E=\frac{J}{\sigma}=\rho J$ ---(1)
Since current is flowing along the axis of the tube so field is also directed along the axis of the tube. Consider the tube to be made up of large number of coaxial annular discs as shown below in the figure

Consider a disc of thickness dx at a distance x from one of the ends of the cylindrical conductor, Thus we have
$E=-\frac{dV}{dx}$
And
$J=\frac{I}{A}=\frac{I}{\pi(b^2-a^2)}$
Where A is the area of cross-section of the cylinder. From equation 1
$\frac{-dV}{dx}=\frac{\rho I}{\pi(b^2-a^2)}$
Or
$\int_{v_1}^{v_2}dV=\int_{0}^{L}\frac{\rho Idx}{\pi(b^2-a^2)}$
Or
$V=V_1-V_2=\frac{\rho LI}{\pi(b^2-a^2)}$
Now Resistance
$R=\frac{V}{I}$
So,
$R=\frac{\rho L}{\pi(b^2-a^2)}$

Question 2
α1 and α2  are the temperature coefficient  of the two resistors R1 and R2 at any  temperature T0 0 C. Find the equivalent temperature coefficient  of their equivalent resistance  if both R1 and R2 are connected in series combination. Assume that  α1 and α2  remain same with change in temperature

At $t= T_0$ °C, equivalent resistance of resistors connected in series is
$R_{eq}= R_1 + R_2$

At t= T °C
$R'_{eq}= R'_1 + R'_2$
$=R_1[1+\alpha_1(T-T_0)]+R2[1+ \alpha _2(T-T_0)]$
$=R_1+R_2+(R_1\alpha_1+R_2\alpha_2)(T-T_0)$ ---(1)
If $\alpha _{eq}$ is equivalent temperature coefficient in series combination then
$R'_{eq}=R_{eq}[1+\alpha_{eq}(T-T_0)]$
$=(R_1+R_2)[1+\alpha_{eq}(T-T_0)]$ ---(2)
From equation 1 and 2,we get
$R_1+R_2+(R_1\alpha_1+R_2\alpha_2)(T-T)= (R_1+R_2)[1+\alpha_{eq}(T-T_0)]$
Solving the equation we get
$\alpha_{eq}=\frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2}$
Again find the $\alpha _{eq}$ for the parallel combination of resistors, Answer shall be
$\alpha_{eq}=\frac{R_1\alpha_2+R_2\alpha_1}{R_1+R_2}$

Question 3
If a copper wire is stretched to make it .2%  longer. Find the percentage change in the resistance

After stretching the wire ,its volume would remain same .If $L_1$ and $A_1$ be the length and area of the wire before stretching and $L_2$ and $A_2$ be the length and area after stretching then
$L_1A_1=L_2A_2=V(Volume)$
Now we know that
$R_1=\frac{\rho L_1}{A_1}=\frac{\rho L_1^2}{V}$
Similarly
$R_2=\frac{\rho L_2}{A_2}=\frac{\rho L_2^2}{V}$
Hence
$R \alpha L^2$
=> $\left(\frac{R_1}{R_2}\right)=\left(\frac{L_1}{L_2}\right)^2$
But As per given data
$L_2= L_1 + \frac{.2 \times L_1}{100} =1.002L_1$
So,
$\left(\frac{R_1}{R_2}\right)=\left(\frac{1}{1.002}\right)^2$
Or
$R_2=(1.002)^2R_1$
Or $R_2=1.004R_1$
Change in resistance =$R_2-R_1 =.004R_1$
% Change in resistance=$\frac{.004R_1 \times 100}{R_1} =.4$ %

Question 4:
An electric current of 5 A passes through a circuit containing three wires(A,B,C) of the same material .Length of the wires in the circuit are in the ratio 2:3:4 And their diameters are in the ratio 3:4:5.Find the amount of current flowing in each branch  of the circuit  when wires are arranged in parallel combination

Let ρ be the resistivity of the material of the wires then resistance of the three wires can be written as
$R_1=\frac{\rho L_1}{A_1}$ ,$R_2=\frac{\rho L_2}{A_2}$ ,$R_3=\frac{\rho L_3}{A_3}$

Also from the given data
L1=2L ,L2=3L,L3=4L
And
D1=3D,D2=4D,D3=5D
Now Areas of the resistors can be written as
$A_1=\pi r_1^2=\pi(\frac{3D}{2})^2$
$A_2=\pi r_2^2=\pi\left(\frac{4D}{2}\right)^2$
$A_3=\pi r_3^2=\pi\left(\frac{5D}{2}\right)^2$

Thus resistance of the wires A,B ,C can be written as
$R_1=\frac{8\rho L}{9\pi D^2}$
$R_2=\frac{3\rho L}{4\pi D^2}$
$R_3=\frac{16\rho L}{25\pi D^2}$
Let I1 , I2 and I3 be the current in the wires A,B ,C respectively. As wires are connected in parallel,

We have
$I_1+I_2+I_3=5$ As these three wires are connected between the same two points  A and B ,we have
V=IReq

Now Req is given by
$\frac{1}{R_{eq}}=\left(\frac{9\pi D^2}{8\rho L}+\frac{4\pi D^2}{3\rho L}+\frac{25\pi D^2}{16\rho L}\right)$
$=\frac{\pi D^2}{\rho L}\left(\frac{9}{8}+\frac{4}{3}+\frac{25}{16}\right)$
$=\frac{193\pi D^2}{48\rho L}$
Or
$R_{eq}=\frac{48\rho L}{193\pi D^2}$
Now
V=IReq
Or
$V=5 \times \frac{48\upsilon L}{193\pi D^2}=\frac{240\rho L}{193\pi D^2}$
Since Potential in Parallel combination is same across all the resistors. Hence current flowing through the first wire
$I_1=\frac{V_1}{R_1}=\left(\frac{240}{193}\right)\left(\frac{\pi D^2}{\rho L}\right)\left(\frac{\rho L}{\pi D^2}\right)\left(\frac{9}{8}\right)=1.4A$
Similarly I2=1.65 A and I3=1.94 A

Question 5:
Find the equivalent resistance between point A and B in following combination of resistors

Let us first label the resistors in the figure

Resistors c and d are connected in series combination
Thus $R_{cd}=R+R=2R$
Now $R_{cd}$ is connected in parallel with resistors e
So equivalent resistance
$\frac{1}{R_{cde}}=\frac{1}{2R}+\frac{1}{2R}$
Or
$R_{cde}=R$
Now the above figure is reduced to below figure

$R_{cde}$ and g are in series
So, $R_{cdeg}=2R$
Now $R_{cdeg}$ and F in parallel
$\frac{1}{R_{AB}}=\frac{1}{2R}+\frac{1}{2R}$
$R_ab=R$ ω

Question 6:
The current density across a cylindrical conductor of Radius A varies according to the following relation

Where K is the constant vector across the length of the conductor and x is the distance from axis.
Find the current through cylindrical cross-section

Let us assume an elemental cylindrical shell of thickness dx at distance x from axis.
Now we know
$J=\frac{I}{A}$
For this elemental cylindrical shell
$J=\frac{dI}{2\pi x}$
Or $dI= K(1-\frac{x}{A})(2\pi x )$
Integrating
$I=\int_{0}^{A}{K\left(1-\frac{x}{A}\right)(}2\pi x)$
$=\frac{K\pi A^2}{3}$

Question 7
Consider the figure given below

Where E=10V is a source of emf with internal resistance r=0.5Ω which is connected to two resistors R1=2Ω and R2=4Ω. Two capacitors C1=0.02µF and C2=0.05µF are connected in parallel to the resistors. Calculate
(a)Current in the circuit and charges Q1 and Q­2 on the capacitors when steady state is reached with switch A open.
(b)Current in the circuit and charges Q1 and Q­2 on the capacitors when steady state is reached with switch A closed

(a) Once the capacitors are fully charged they do not draw any current from the battery. Thus in a steady state
$I=\frac{\xi}{r+R_1+R_2}=\frac{10}{.5+2+4}=1.53$A
The potential difference across the capacitor is
$V=I\left(R_1+R_2\right)=1.53\left(2+4\right)=9.18$V
Since the capacitors are connected in series so, the equivalent capacitance is
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$
Using the values of $C_1$ and $C_2$ in above equation we find
$C_{eq}=0.0143$ μ F
Charge on a capacitor with capacitance equal to equivalent capacitance is $Q=C_{eq}V$
Thus
$Q=C_{eq}V=9.18 \times 0.0143=0.131$ μ C
And
$Q=Q_1=Q_2=0.131$ μ C
(b) When the switch A is closed the steady state current remains the same that is I=1.53A but capacitors are no longer connected in series combination so,
$V_1=IR_1=1.53 \times 2=3.06$V
$V_2=IR_2=1.53 \times 4=6.12$V
And charge on $C_1$ equals
$Q_1=C_1 V_1=0.02 \times 3.06=.0612$ μ C
$Q_2=C_2V_2=0.05 \times 6.12=.306$ μ C

Question 8
Consider the infinite network of resistors as shown below in the figure

Each resistor in the infinite network has a value of 2Ω. Find the current drawn from 12V supply of internal resistance 0.5Ω.

Let X is the equivalent resistance of the entire network. Since the network is infinite, so adding more set of resistance each of resistance R=2 ω across the terminals should not affect total resistance i.e., it should still remain equal to X. Figure given below shows the network after adding one set of three resistances.
If R' is equivalent resistance of this new network then,
R'=R+(resistance equivalent to parallel combination of X and R) + R
$=2R+\frac{XR}{X+R}$
Now since addition of one set of three resistances should not alter the total resistance of the infinite network. So we have
$R'=X$
Or,
$2R+\frac{XR}{X+R}=X$
Since, R=2 ω
$4+\frac{2X}{X+2}=X$
$4X+8+2X=X^2+2X$
$X^2-4X-8=0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$X=\frac{-(-4)\pm\sqrt{16-4(-8)}}{2}$
$X=\frac{4\pm4\ \sqrt3}{2}$
Ignoring X with negative value we find
X=5.46 ω
Now E=12V and r=.5 ω
Current drawn by the network is
$I=\frac{E}{X+r}$
$I=\frac {12}{5.46+.5}$
I=2.01A

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