Current Electricity

Case 1 : When $\mathrm{S}_1$ and $\mathrm{S}_2$ are open

$ \left(R_{e q}\right)_1=1+\frac{5 \times \frac{1}{2}}{5+\frac{1}{2}}=1+\frac{5}{11}=\frac{16}{11} $

$ P_1=\frac{V^2}{(R_{e q})_1}=\frac{(6)^2}{16} \times 11=\frac{36 \times 11}{16}=24.75 \mathrm{~W} $

$ \left(R_{e q}\right)_2=\frac{6}{11} \Omega $

$ P_2=\frac{V^2}{(R_{e q})_2}=\frac{(6)^2}{6} \times 11=\frac{36 \times 11}{6}=66 \mathrm{~W} $

$ P_2>P_1 $

Option (A) is correct.

Case 2 : If I = $2 \mathrm{~A}$ source is used in both circuits, then

$ P_1=i^2\left(R_{e q}\right)_1=(2)^2 \times \frac{16}{11}=\frac{64}{11}=5.818 \mathrm{~W} $

$ P_2=i^2\left(R_{e q}\right)_2=(2)^2 \times \frac{6}{11}=\frac{24}{11}=2.1818 \mathrm{~W} $

$ P_1>P_2 $

Option (B) is correct.

Case 3 : For $Q_1$

$ \begin{aligned} & R_{e q}=\frac{5}{11} \Omega \\\\ & Q_1=\frac{V^2}{R_{e q}}=\frac{(6)^2}{\frac{5}{11}}=\frac{36 \times 11}{5}=79.2 \mathrm{~W} \\\\ & P_1=24.75 \mathrm{~W} \\\\ & Q_1>P_1 \end{aligned} $

Option (C) is correct.

Case 4 : For option (D)

$ \begin{aligned} & Q_1=i^2 R_{e q}=(2)^2 \times \frac{5}{11}=\frac{20}{11}=1.81 \mathrm{~W} \\\\ & Q_2=i^2 R_{e q}=(2)^2 \times \frac{1}{2}=\frac{4}{2}=2 \mathrm{~W} \\\\ & Q_2>Q_1 \end{aligned} $

Option (D) is incorrect.

Point A and B are at same potential so they can be merged/folded.

For loop 1

$ \begin{aligned} & -\frac{1}{2} x-\frac{1}{2}(x+y)-x+6=0 \\\\ &\Rightarrow -2 x-\frac{y}{2}=-6 \\\\ &\Rightarrow -4 x-y=-12 \\\\ &\Rightarrow 4 x+y=12 .......(1) \end{aligned} $

For loop 2

$ \begin{aligned} & \frac{1}{2} y+1 y+12+\frac{1}{2}(x+y)=0 \\\\ &\Rightarrow \frac{3}{2} y+\frac{y}{2}+\frac{x}{2}=-12 \\\\ &\Rightarrow 2 y+\frac{x}{2}=-12 \\\\ & \begin{array}{l} \Rightarrow4 y+x=-24 \\\\ \Rightarrow4 y+x-16 x-4 y =-24-48 \end{array} \\\\ &\Rightarrow -15 x=-72 \end{aligned} $

$ \Rightarrow $ $x=\frac{72}{15}$

$ \Rightarrow $ $4\left(\frac{72}{15}\right)+y=12$

$ \begin{aligned} \Rightarrow y & =12-\frac{288}{15} \\\\ & =\frac{180-288}{15} \\\\ & =\frac{-108}{15}=-7.2 \mathrm{~A} \end{aligned} $

Current in $R_1=7.2 \mathrm{~A}$

Current in $R_2=\frac{x+y}{2}=\left(\frac{72}{15}-\frac{108}{15}\right) \frac{1}{2}$

$ \begin{aligned} & =\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2} \mathrm{~A} \\\\ & =1.2 \mathrm{~A} \end{aligned} $

Current in $R_3=x=4.8 \mathrm{~A}$

Current in $R_5=\frac{1}{2} x=2.4 \mathrm{~A}$

All the elements are in parallel.

$ \therefore $ $\int {{l \over {dr}} = \int\limits_{{R_1}}^{{R_2}} {{{t\,dx} \over {\rho \,\pi x}}} } $

${l \over r} = {t \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$

Resistance = ${{\pi \rho } \over {t\ln \left( {{{{R_2}} \over {{R_1}}}} \right)}}$

$ \therefore $ $I = {V \over {{\mathop{\rm Re}\nolimits} sis\tan ce}}$

$ = {{{V_t}} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$



$eE = {{m{v^2}} \over r}$

$I = neA{V_d}$

${{{V_0}tdr} \over {\rho \pi r}} = ne(drt)V$

$ \Rightarrow V = \sqrt {{{{V_0}} \over {\rho \pi ner}}} $

$ \therefore $ $E = {m \over {er}}{{V_0^2{t^2}} \over {{\rho ^2}{\pi ^2}{n^2}{e^2}{r^2}}}$

$ \Rightarrow E = {{mV_0^2} \over {{e^3}{r^2}{\rho ^2}{\pi ^2}{n^2}}}$

$dV = - Edr \Rightarrow dV = - {K \over {{r^3}}}dr$

$V = K\int\limits_{{R^1}}^{{R_2}} {{r^{ - 3}}dr} $

$\Delta V = {{mV_0^2} \over {{e^3}{\rho ^2}{\pi ^2}{n^2}}}\left( {{1 \over {r_1^2}} - {1 \over {r_2^2}}} \right)$

$\Delta V \propto V_0^2 $

$ \Rightarrow $ $\Delta $V $ \propto $ I² ( As $ I \propto {V_0}$ )

Just after closing of switch charge on any capacitor is zero.

$ \therefore $ Replace all capacitors by conducting wires.



Current flow in the circuit,

$i = {5 \over {70 + 100 + 30}} = {5 \over {200}} = 25$ mA

Now S₁ is kept closed for long time circuit is in steady state.



${q \over {10}} + {q \over {80}} + {q \over {80}} - 5 = 0 \Rightarrow {{10q} \over {80}} = 5$

$ \therefore $ $q = 40\mu C$

$ \therefore $ V across ${C_1} = {{40} \over {10}} = 4V$

Now just after closing S₂ charge on each capacitor remains same.



Applying KVL,

$ - 10 + x \times 30 + {{40} \over {10}} + y \times 70 = 0$

30x + 70y = 6 .....(1)

$ - {{40} \over {80}} + 5 + (x - y)30 - {{40} \over {80}} + (x + y) \times 100 - 10 + x \times 30 = 0$

$160x - 130y - 6 = 0$ .... (2)

$y = {{96} \over {1510}}$

x = 0.05 A

$V = 100 \times {10^{ - 3}} = {10^{ - 1}}V$

$V = {{\mathop{\rm l}\nolimits} _g}({R_g} + {R_v})$

${{{{10}^{ - 1}}} \over {2 \times {{10}^{ - 6}}}}={R_g} + {R_v}$

$5 \times {10^4}\Omega \approx {R_v}$ ($ \because $${R_v} < {10^5}\Omega $)



I_gR_g = (I - I_g)S


$S = {{2 \times {{10}^{ - 6}} \times 10} \over {{{10}^{ - 3}} - 2 \times {{10}^{ - 6}}}}$

$S = 2 \times {10^{ - 5}} \times {10^3} = 2 \times {10^{ - 2}} = 20m\Omega $

${R_A} = {{S{R_g}} \over {S + {R_g}}} = {{20 \times {{10}^{ - 3}} \times 10} \over {10 + 20 \times {{10}^{ - 3}}}}$

$ \approx 20 \times {10^{ - 3}}\Omega $

$i = {\varepsilon \over {\left( {{{1000 \times 50 \times {{10}^3}} \over {51 \times {{10}^3}}}} \right)}} = {{51\varepsilon } \over {5 \times {{10}^4}}}$

($ \because $ R_A $ \to $ 0)

$i' = i\left( {{{{R_v}} \over {51 \times {{10}^3}}}} \right) = {\varepsilon \over {1000}}$

Measured resistance,

$ \therefore $ ${R_m} = {{i' \times 1000} \over i} = {\varepsilon \over {51\varepsilon }} \times 5 \times 10 = 980.4\Omega $

If the voltmeter shows full scale deflection then

${\varepsilon \over {980}} \times \left( {{{1000} \over {51 \times {{10}^3}}}} \right) \times 5 \times {10^4} = {10^{ - 1}}$

$\varepsilon = 999.6$ mV

Since, i_A = 1 mA so maximum reading of R can be

${{999.6\,mV} \over {1\,mA}} = 999.6\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 i₁ and i₂ be the currents as shown in the figure.


Apply Kirchhoff's law in the loop ABCDA and CEFDC to get

${i_1}{R_1} + ({i_1} - {i_2}){R_2} = {V_1}$ ..... (1)

${i_2}{R_3} - ({i_1} - {i_2}){R_2} = {V_2}$ ...... (2)

Multiply equation (1) by R₃ and (2) by R₁ and then subtract to get the current through R₂ as

$({i_1} - {i_2}) = {{{V_1}{R_3} - {V_2}{R_1}} \over {{R_1}{R_3} + {R_2}{R_3} - {R_1}{R_2}}}$.

The current through R₂ becomes zero when V₁R₃ = V₂R₁.

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, ${V_S} < {V_Q}$

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 ${(i')^2}$:

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

Given, current carrying circular coil does not have an electric field and a gravitational field.

But it has a magnetic field and induced electric field if the magnetic field due to current in the coil is changing with time.