Current Electricity

Given : Volume charge density of rod = $\rho$.

We know that current $I = neA{v_d}$; where n is number of electrons, e is electronic charge, A is area, v_d is drift velocity

Since volume charge density is $\rho$ = ne. Therefore,

$I = \rho A{v_d} \Rightarrow {v_d} = {I \over {{\rho _A}}}$

Now, time = ${{dis\tan ce} \over {speed}}$

$\Rightarrow$ time $ = {d \over {{v_d}}} = {d \over {I/\rho A}} = {{\rho Ad} \over I}$

Thus, time required to travel distance $d = {{\rho dA} \over I}$

Since rate of heat $ = {{{V^2}} \over R} \Rightarrow $ rate of heat $ \propto {1 \over R}$, where $R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}$; here r is radius of wire and L is length of wire. Therefore,

$R \propto {L \over {{r^2}}}$.

Thus, rate of heat $ \propto {{{r^2}} \over L}$.

When length in halved and radius is doubled then, rate of heat $ \propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}$

Therefore, rate of heat developed in wire will be increased 8 times.

We have the following cases :

$\bullet$ For configuration (I) : Considering as Wheatstone's bridge, we get the equivalent resistance as ${({R_I})_{eq}} = 1\,\Omega $.

$\bullet$ For configuration (II) : The given configuration is further reduced as follows:

That is, ${({R_{II}})_{eq}} = {1 \over 2}\,\Omega $.

$\bullet$ For configuration (III) : Similarly, we get ${({R_{III}})_{eq}} = 2\,\Omega $.

The power is given by

$P = {{{V^2}} \over R}$

Since V is the same for all three configurations, we have the power as a quantity indirectly proportional to the resistance :

$P \propto {1 \over R}$

From the above three case, we have ${({R_I})_{eq}} = 1\,\Omega ;\,{({R_{II}})_{eq}} = {1 \over 2}\,\Omega ;\,{({R_{III}})_{eq}} = 2\,\Omega $, which implies that

${R_{II}} > {R_I} > {R_{III}}$

Therefore, ${P_2} > {P_2} > {P_3}$

Initially, the resistance P = 4 $\Omega$ and the neutral point N is at 60 cm from A. Therefore,

${4 \over {60}} = {Q \over {40}} \Rightarrow Q = {{160} \over {60}} = \left( {{8 \over 3}} \right)\Omega $

When an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A, we have

${{4 + R} \over {80}} = {Q \over {20}}$

$ \Rightarrow {{4 + R} \over {80}} = {8 \over {60}} \Rightarrow 4 + R = {{64} \over 6}$

$ \Rightarrow R = {{64} \over 6} - 4 = {{64 - 24} \over 6} = {{40} \over 6}$

$ \Rightarrow R = \left( {{{20} \over 3}} \right)\Omega $

Let i_g be the current that gives full deflection of the galvanometer. When all components are connected in series (see figure), effective resistance of the circuit is ${R_e} = 2R + 2{R_c}$ and maximum current allowed in the circuit is i_g.

Thus, the voltage between A and B is

${V_{AB}} = {i_g}{R_e} = 2{i_g}(R + {R_c})$.

Consider the case when two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel.

The maximum current through each galvanometer is i_g and the maximum current through the resistors is 2i_g. Apply Kirchhoff's law to get the voltage between A and B as

$V{'_{AB}} = 2{i_g}R + 2{i_g}R + {i_g}{R_c}$

$ = 2{i_g}(R + {R_c}) + 2{i_g}(R - {R_c}/2)$

$ = {V_{AB}} + 2{i_g}(R - {R_c}/2)$

$ > {V_{AB}}$ ($\because$ ${R_c} < R/2$).

Consider the case when all four components are connected in parallel.

Let i and i_g be the currents through the resistors and the galvanometers. By Kirchhoff's law, $iR = {i_g}{R_c}$, which gives $i = {i_g}{R_c}/R$. The current between A and B is

${I_{AB}} = 2i + 2{i_g} = 2{i_g}(1 + {R_c}/R)$.

Consider the case when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

$I{'_{AB}} = {i_g} + 2i = {i_g} + 2{i_g}(2{R_c}/R)$

$ = 2{i_g}(1 + {R_c}/R) - (1 - 2{R_c}/R){i_g}$

$ = {I_{AB}} - (1 - 2{R_c}/R){i_g}$

$ < {I_{AB}}$ ($\because$ ${R_c} < R/2$).

When the key is just pressed, there is no charge on capacitors

$\therefore$ $5 = 25000{I_1} \Rightarrow {I_1} = {5 \over {25000}} = 0.2\,mA$

and $5 = 50000{I_2} \Rightarrow {I_2} = {5 \over {50000}} = 0.1\,mA$

Reading of voltmeter $ = {V_Q} - {V_P} = - 25000{I_1}$

$ = - 25000 \times 0.2 \times {10^{ - 3}}V = - 5V$

After a very long time, steady state is reached, i.e. ${I_1} - {I_2} = 0$

$\therefore$ $5 = {{{q_1}} \over {40 \times {{10}^{ - 6}}}}$ or ${q_1} = 200\mu C$

and $\therefore$ $5 = {{{q_2}} \over {20 \times {{10}^{ - 6}}}}$ or ${q_2} = 100\mu C$

Now, reading of voltmeter $ = {V_Q} - {V_P}$

$ = {{{q_2}} \over {20\mu F}} = {{100\mu C} \over {20\mu F}} = 5V$

For capacitor of 40$\mu$F,

${\tau _1} = RC = 25 \times {10^3} \times 40 \times {10^{ - 6}} = 1\,s$

and for capacitor of 20$\mu$F,

${\tau _2} = RC = 50 \times {10^3} \times 20 \times {10^{ - 6}} = 1\,s$

$\therefore$ At any instant t,

${q_1} = 200[1 - {e^{ - t}}]\mu C$, ${q_2} = 100[1 - {e^{ - t}}]\mu C$

${I_1} = 0.2\,{e^{ - t}}$ mA, ${I_2} = 0.1\,{e^{ - t}}$ mA

${V_Q} - {V_P} = {{100(1 - {e^{ - t}})\mu C} \over {20\mu F}} - 25k\Omega \times 0.2\,{e^{ - t}}$ mA

$ = 5(1 - {e^{ - t}}) - 5{e^{ - t}} = 5 - 10{e^{ - t}}$

At t = ln 2 s; ${V_Q} - {V_P} = 5 - 10{e^{ - \ln 2}} = 5 - {{10} \over 2} = 0$ V

Initially, I₀ = 0.1 mA + 0.2 mA = 0.3 mA

At t = 1 s, I = I₁ + I₂

$ = (0.2{e^{ - 1}} + 0.1{e^{ - 1}}) = {{0.3} \over e}\,mA = {{{I_0}} \over e}$

After a long time, i.e. t $\to$ $\infty$

$I = {I_1} + {I_2} = 0.2{e^{ - \infty }} + 0.1{e^{ - \infty }} = 0$

Because of non-uniform evaporation at different section, area of cross-section would be different at different sections.

Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section.

Resistance of the wire as whole increases with time. Overall resistance increases hence power decreases.

($p = {{{V^2}} \over R}$ or $p \propto {1 \over R}$ as V is constant).

At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

Let the charge per unit length on the axis be $\lambda$(t) at a time t. The electric field due to this line charge at a radial distance r is given by

$\overrightarrow E (r,t) = {{\lambda (t)} \over {2\pi \in r}}\widehat r$. ...... (1)

The free charges inside the conductor start moving radially due to the presence of electric field. The current density at a point is defined as the current flowing across a unit area placed perpendicular to the direction of current flow. The current density at a point is related to the electric field at that point by Ohm's law i.e.,

$\overrightarrow j (r,t) = \sigma \overrightarrow E (r,t)$, ....... (2)

where $\sigma$ is the electrical conductivity of the conductor. The current through a cylindrical shell of radius r and length l is given by

$I = \int_{surface} {\overrightarrow j (r,t)\,.\,d\overrightarrow A = j(r,t)(2\pi rl)} $. ...... (3)

By conservation of charge, the current I is equal to the rate of decrease of charge q on axial line segment of length l i.e.,

$I = - {{dq} \over {dt}} = - l{{d\lambda (t)} \over {dt}}$. ...... (4)

From equations (1) - (4)

${{d\lambda (t)} \over {dt}} = - {{\sigma \lambda (t)} \over \in }$.

Integrate with initial condition $\lambda$(t = 0) = $\lambda$₀ to get

$\lambda (t) = {\lambda _0}\exp {e^{ - {{\sigma t} \over \in }}}$,

Substitute $\lambda$(t) in equation (2) to get

$\overrightarrow j (r,t) = {{\sigma {\lambda _0}} \over {2\pi \in r}}{e^{ - {{\sigma t} \over \in }}}$.

Note that current density varies as 1/r with radial distance r. It decreases exponentially with time and becomes zero at t $\to$ $\infty$.

Resistance of a wire, $R = {{\rho l} \over A}$

For iron (Fe) bar,

$\rho$ = 10^$-$7 $\Omega$m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (2 mm) $\times$ (2 mm) = 4 mm² = 4 $\times$ 10^$-$6 m²

${R_1} = {{{{10}^{ - 7}} \times 50 \times {{10}^{ - 3}}} \over {4 \times {{10}^{ - 6}}}} = 1250 \times {10^{ - 6}}$ $\Omega$ = 1250 $\mu$W

For aluminium (Al) bar,

$\rho$ = 2.7 $\times$ 10^$-$8 $\Omega$ m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (7² $-$ 2²) mm² = 45 mm² = 45 $\times$ 10^$-$6 m²

$\therefore$ ${R_2} = {{2.7 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 3}}} \over {45 \times {{10}^{ - 6}}}}$

$ = {{27 \times 50} \over {45}} \times {10^{ - 6}} = 30 \times {10^{ - 6}}$ $\Omega$ = 30 $\mu$$\Omega$

Potential difference across both bars (resistors) is same so they are in parallel combination.

Equivalent resistance between P and Q is given by

$R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 \times 30} \over {1250 + 30}} = {{125 \times 30} \over {128}} = {{1875} \over {64}}\mu \Omega $.

when a metre bridge is balanced, then

${{{R_1}} \over x} = {{{R_2}} \over {(100 - x)}}$ ..... (i)

From figure,

R₁ = R, R₂ = 90 $\Omega$, x = 40 cm

Then, ${R \over {40}} = {{90} \over {(100 - 40)}} = {{90} \over {60}}$

$\Rightarrow$ R = 60 $\Omega$

Now, for $\Delta$R taking natural log on both sides of eqn. (i),

$\ln {R_1} = \ln {R_2} + \ln x - \ln (100 - x)$

or, $\ln R = \ln x - \ln (100 - x) + \ln (90)$

On differentiating,

${{\Delta R} \over R} = {{\Delta x} \over x} - {{\Delta (100 - x)} \over {(100 - x)}}$

$ \Rightarrow {{\Delta R} \over R} = {{\Delta x} \over x} + {{\Delta x} \over {(100 - x)}}$

$ \Rightarrow \Delta R = \left( {{{0.1} \over {40}} + {{0.1} \over {60}}} \right) \times 60 = {{0.5} \over {120}} \times 60 = 0.25\Omega $

$\therefore$ Required value of R = (60 $\pm$ 0.25) $\Omega$

P = Vi

$\therefore$ $i = {P \over V} = {{600 \times {{10}^3}} \over {4000}} = 150$ A

Total resistance of cables,

R = 0.4 $\times$ 20 = 8$\Omega$

$\therefore$ Power loss in cables = i²R

= (150)² (8)

= 180000 W = 180 kW

This loss is 30% of 600 kW.

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown below :

${I_2} = {{12} \over {2 + 2 + 2}} = 2A$

${I_3} = {{12} \over {4 + 4 + 4}} = 1A$

$\therefore$ ${I_1} = {I_2} + {I_3} = 3A$

I_PQ = 0 because V_P = V_Q

Potential drop (from left to right) across each resistance is

${{12} \over 3} = 4V$

$\therefore$ ${V_{MS}} = 2 \times 4 = 8V$

${V_{NQ}} = 1 \times 4 = 4V$

or, ${V_S} < {V_Q}$

Corrected length L₁ (AJ) = 52 + 1 = 53 cm

Corrected length L₂ (BJ) = (100 $-$ 52) + 2 = 50 cm

For a balanced Wheatstone bridge,

${X \over {10}} = {{{L_1}} \over {{L_2}}} = {{53} \over {50}}$

$\Rightarrow$ X = 10.6 $\Omega$

The power of the bulb is

$P = {{{V^2}} \over R}$

Therefore,

$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$

where R₁₀₀ is the resistance (at any temperature) corresponds to 100 W. Similarly,

$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$ and $40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$

From these equations, we get

${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$

To verify Ohm's law, we need to measure the voltage across the test resistance R_T and current passing through it. The voltage can be measured by connecting a high resistance R₁ in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to R_T. The current can be measured by connecting a low resistance R₂ (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through R_T.

The sheet is of square shape with thickness t, width $\omega$ = L, and length l = L. The resistance between the two opposite faces is given by

$R = {{\rho l} \over A} = {{\rho l} \over {wt}} = {{\rho L} \over {Lt}} = {\rho \over t}$.

The current in the net circuit is

$I = {V \over {{R_{effective}}}}$

$ = {V \over {{R_1} + [({R_2}{R_L})/({R_1} + {R_L})]}}$

$ = {{24} \over {2 \times {{10}^{ - 3}} + [(6 \times 1.5 \times {{10}^{ - 6}})/(7.5 \times {{10}^{ - 3}})]}} = 7.5$ mA

The potential across $R_L$ is

${V_L} = I \times {{{R_2}{R_L}} \over {{R_2} + {R_L}}}$

$ = 7.5 \times {10^{ - 3}} \times 1.2 \times {10^3} = 9$ V

Therefore,

$I = {9 \over {{R_L}}} = 6$ mA

The ratio of power dissipated in $R_1$ and $R_2$ is

${{I{R_1}} \over {(I - i){R_2}}} = {{{{(7.5 \times {{10}^{ - 3}})}^2} \times 2 \times {{10}^3}} \over {{{(1.5 \times {{10}^{ - 3}})}^2}6 \times {{10}^3}}} = 0.75$ J

The power dissipated in $R_L$ is

${{V_L^2} \over {{R_L}}} = {{{9^2}} \over {1.5 \times {{10}^3}}} = 54$ mJ

If $R_1$ and $R_2$ are interchanged, then

$I' = {V \over {{R_{effective}}}}$

$ = {V \over {{R_2} + [({R_1}{R_L})/({R_L} + {R_1})]}}$

$ = {{24} \over {6 \times {{10}^{ - 3}} + [(2 \times 1.5 \times {{10}^{ - 6}})/3.5 \times {{10}^{ - 3}}]}} = 3.5$ mA

Hence, the voltage drop across the load is

$V{'_L} = I \times {{{R_1}{R_L}} \over {{R_1} + {R_L}}} = 1$ V

Therefore, $i' = 2$ mA. The magnitude of the power dissipated is the ratio between ${i^2}$ and ${(i')^2}$:

${{{P_1}} \over {{P_2}}} = {{{i^2}} \over {{{(i')}^2}}} = {{{6^2}} \over {{2^2}}} = 9$

We know that $P = {{{V^2}} \over R}$

If potential is constant we have

$P \propto {1 \over R}$

Case I

Hence, this is a clear case of wheat stone bridge R$_1$ = 1 $\Omega$

Case II

Equivalent resistance (R$_2$)

$\therefore$ ${1 \over {{R_2}}} = {1 \over 2} + {1 \over 1} + {1 \over 2} = {{1 + 2 + 1} \over 2} = {4 \over 2} = 2\,\Omega $

Hence, ${R_2} = {1 \over 2} = 0.5\,\Omega $

It is clear that the equivalent resistance (R$_2$) will be less than 1 $\Omega$.

Case III

Hence, R$_3$ = 2 $\Omega$

Since, R$_2$ < R$_1$ < R$_2$

$\therefore$ P$_2$ > P$_1$ > P$_3$ ($ \because $ $P \propto {1 \over R}$)

When the temperature of a metal increases, its resistance will increase.

Therefore statement-2 is correct.

For meter bridge, when null point N is obtained, we get

${X \over l} = {R \over {100 - l}}$

When the unknown resistance is put inside an enclosure, maintained at a high temperature, then X increases. To maintain the ratio of null point, $l$ should also increase. But if we want to keep the null point at the initial position (i.e., if we want to change the value of $l$) to maintain the ratio, R should be increased.

Therefore, statement - 1 is false.