Current Electricity

Given, Length of wire $L=5 \mathrm{~m}$

Resistance of wire $=5 \Omega=R$

Balancing point $=3 \mathrm{~m}$

Emf of cell $=10 V=E_1$

Current in circuit with $E_1=10 \mathrm{~V}$ is circuit with $E_1=10 \mathrm{~V}$ is given by

$ \begin{aligned} I & =\frac{E_1}{R+r}[r=4+1 \Omega=5 \Omega] \\ I & =\frac{10}{5+5}=1 \mathrm{~A} \\ V_{A B} & =I \times R=1 \times 5=5 \mathrm{~V} \end{aligned} $

Potential gradient is given by

$ \begin{aligned} & K=\frac{V_{A B}}{l(\text { for whole wire })} \\ & K=5 / 5=1 \end{aligned} $

Now, at balancing point.

$ \begin{aligned} & E=k l \Rightarrow E_1=1 \times 3 \\ & E_1=3 \mathrm{~V} \end{aligned} $

Resolving the circuit

$10 \Omega$ and $10 \Omega$ are in parallei connect and 6 and $6 \Omega$ are in parallel connection.

$ \begin{aligned} \because \text { New resistance } & =\frac{10 \times 10}{10+10}=5 \Omega \\ & =\frac{6 \times 6}{6+6}=3 \Omega \end{aligned} $

Redrawing the circuit diagram.

The ratio of resistance on both side of junction is $1: 1$, this condition is equal to wheat stone bridge condition. No, current will passes through $8 \Omega$.

New circuit will be equivalent resistance $=R_{\mathrm{eq}}=(5+3)=8 \Omega$ each side They are in parallel connection

$ R_{\mathrm{eq}}=\frac{8 \times 8}{8+8}=4 \Omega $

Given,

Wheat stone bridge is balanced.

Power dissipated through $R_3=P_3$

Power dissipated through $R_1=P_1$

As, wheat stone bridge is balanced no current will flow from galvanometer arm. The condition of balancing of bridge is

$ \frac{R_1}{R_3}=\frac{L}{K} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Power discipated by resistance circuit is given by,

$ \begin{aligned} & P=V I \\or\,\, & P=I^2 R \end{aligned} $

Power dissipation through $R_1$ and $R_3$ are

$ P_3=I^2 R_3, P_1=I^2 R_1 $

Ratio of power discipation.

$ \frac{P_3}{P_1}=\frac{I^2 R_3}{I^2 R_1} \Rightarrow \frac{P_3}{P_1}=\frac{R_3}{R_1}=\frac{K}{L} $

Given,

Resistance of wire $=2 R$

Let, Radius of wire $=r$

Length of wire $=L$

Asea of wire $=A$

When the length of wire is stretched to double volume will remains same.

$ \begin{aligned} V_{\text {intial }} & =V_{\text {final }} \\ L \times A & =L^{\prime} A^{\prime} \\ A^{\prime} & =\frac{L \times A}{2 L}=\frac{A}{2} \end{aligned} $

Initial resistance

$ R_1=2 R=\frac{\rho L}{A^{\prime}} \,\,\,\,\,\,\,[Given]$

New resistance

$ \begin{aligned} & R_2=\frac{\rho L^{\prime}}{A^{\prime}} \Rightarrow \frac{\rho 2 L}{\frac{A}{2}} \Rightarrow R_2=\frac{4 \rho L}{A}=4 \times 2 R \\ & R_2=8 R \end{aligned} $

So, the increase in resistance is $R^{\prime}-R$

$ 8 R-2 R=6 R $

${ }_{}$Given, $m_1: m_2: m_3=1: 2: 3$

$ \begin{aligned} & \therefore m_1=x, m_2=2 x \text { and } m_3=3 x \\ & \text { and } l_1: l_2: l_3=3: 2: 1 \end{aligned} $

$l_1=3 K, l_2=2 K$ and $l_3=K$

Resistance of the wire is given by

$ \begin{aligned} R & =\rho \frac{l}{A}=\rho \cdot \frac{l^2}{l \cdot A}=\rho \cdot \frac{l^2}{V} \\ & =\frac{\rho l^2}{\left(\frac{m}{d}\right)}=\frac{\rho l^2 d}{m} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad R \propto \frac{l^2}{m} \Rightarrow R_1 \propto \frac{l_1^2}{m_1} \\ & \Rightarrow \quad R_1 \propto \frac{(3 K)^2}{x} \\ & \Rightarrow \quad R_1 \propto \frac{9 K^2}{x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Similarly, } R_2 \alpha \frac{l_2^2}{m_2} \Rightarrow R_2 \propto \frac{(2 K)^2}{2 x} \\ & \Rightarrow \quad R_2 \propto \frac{(2 K)^2}{2 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } \quad R_3 \propto \frac{l_3^2}{m_3} \\ & \Rightarrow \quad R_3 \propto \frac{K^2}{3 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { From Eqs. (i), (ii) and (iii), we get }\\ &R_1: R_2: R_3=9: 2: \frac{1}{3}=27: 6: 1 \end{aligned} $

$ \begin{aligned} &\text { Wheatstone's bridge is balanced if }\\ &\frac{P}{Q}=\frac{R}{S_1 \| S_2} \end{aligned} $

$ \Rightarrow \frac{P}{Q}=\frac{P}{\left(\frac{S_1 S_2}{S_1+S_2}\right)} \Rightarrow \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2} $

The resistance of a wire, $R$, depends on the material, the length ($l$), and the cross-sectional area ($A$). The formula is:

$ R = \rho \cdot \frac{l}{A} $ where $\rho$ is the resistivity (which depends on the material).

The area $A$ of a wire with radius $r$ is $A = \pi r^2$. So the resistance can be rewritten as: $ R = \rho \cdot \frac{l}{\pi r^2} $ This means that resistance goes up if the length ($l$) increases, and goes down if the radius ($r$) increases.

If the length is doubled ($l_2 = 2l_1$), and the radius is also doubled ($r_2 = 2r_1$), we can find the new resistance $R_2$ compared to the old resistance $R_1 = R$.

When both length and radius change, the formula for comparing the new and old resistance becomes: $ \frac{R_2}{R_1} = \left( \frac{r_1}{r_2} \right)^2 \cdot \frac{l_2}{l_1} $

Plug in the values ($r_2 = 2r_1$, $l_2 = 2l_1$): $ \frac{R_2}{R} = \left( \frac{r_1}{2r_1} \right)^2 \cdot \frac{2l_1}{l_1} = \left( \frac{1}{2} \right)^2 \times 2 = \frac{1}{4} \times 2 = \frac{1}{2} $

This means the new resistance is half the original: $ R_2 = \frac{R}{2} $

The specific resistance (or resistivity, $\rho$) does not change when you change the size or length of the material. It only depends on what the wire is made of and its temperature.

$ \begin{aligned} &\text { We know that, in meter bridge }\\ &\begin{aligned} & & \frac{R_{\text {left }}}{R_{\text {right }}} & =\frac{l}{100-l} \\ & \Rightarrow & \frac{2}{3} & =\frac{l}{100-l} \\ & \Rightarrow & 200-2 l & =3 l \Rightarrow 200=5 l \\ & \Rightarrow & 40 & =l \Rightarrow l=40 \mathrm{~cm} \end{aligned} \end{aligned} $

According to given condition,

$ \begin{array}{ll} & I_g=5 \% \text { of } I=0.05 I \\ \therefore & I-I_g=I-0.05 I=0.95 I \end{array} $

Applying KVL in the loop, we get

$ \begin{aligned} & \left(I-I_g\right) S-I_g G=0 \\ \Rightarrow \quad & \left(I-I_g\right) S=I_g G \\ \Rightarrow \quad & (0.95 I) S=(0.05 I) G \Rightarrow S=\frac{G}{19} \end{aligned} $

Given:

Internal resistance of the cell, $ r = 1 \, \Omega $

Initial balancing length, $ l_1 = 44 \, \text{cm} $

New balancing length, $ l_2 = 40 \, \text{cm} $

Formula:

The internal resistance of the cell can be expressed as:

$ r = R \left(\frac{l_1}{l_2} - 1\right) $

Here, $ R $ is the external resistance connected parallel to the cell. We know $ r = 1 \, \Omega $.

Calculation:

$ 1 = R \left(\frac{44}{40} - 1\right) $

Simplify the expression:

$ 1 = R \left(\frac{44}{40} - \frac{40}{40}\right) $

$ 1 = R \left(\frac{4}{40}\right) $

$ 1 = \frac{R}{10} $

Solve for $ R $:

$ R = 10 \, \Omega $

Thus, the resistance that needs to be connected in parallel to the cell to achieve the desired balancing point is $ 10 \, \Omega $.

Given the problem, two wires made of the same material have lengths and radii in specified ratios, and the same potential difference is applied to both. Our goal is to find the ratio of electric currents through the wires.

Given values:

Ratio of lengths of the wires: $\frac{L_1}{L_2} = \frac{2}{3}$

Ratio of radii of the wires: $\frac{r_1}{r_2} = \frac{8}{9}$

Since both wires are made of the same material and experience the same potential difference, we use Ohm's Law:

$ V = I R $

This leads to the relationship for the wires:

For the first wire: $V = I_1 R_1$

For the second wire: $V = I_2 R_2$

Equating the two since $V$ is the same:

$ I_1 R_1 = I_2 R_2 $

This implies:

$ \frac{I_1}{I_2} = \frac{R_2}{R_1} $

The resistance $R$ is given by:

$ R = \rho \frac{L}{A} $

where:

$\rho$ is the resistivity,

$L$ is the length,

$A = \pi r^2$ is the cross-sectional area.

Thus, for the resistances:

$ R_1 = \rho \frac{L_1}{\pi (r_1)^2}, \quad R_2 = \rho \frac{L_2}{\pi (r_2)^2} $

Substituting these into the current ratio equation:

$ \frac{I_1}{I_2} = \frac{\rho \frac{L_2}{\pi (r_2)^2}}{\rho \frac{L_1}{\pi (r_1)^2}} = \frac{L_2}{L_1} \times \left(\frac{r_1}{r_2}\right)^2 $

Substitute the given ratios:

$ \frac{I_1}{I_2} = \left(\frac{3}{2}\right) \times \left(\frac{8}{9}\right)^2 = \frac{3}{2} \times \frac{64}{81} = \frac{192}{162} = \frac{32}{27} $

Thus, the ratio of current $I_1 : I_2$ is $32 : 27$.

When the potentiometer is connected between points $ A $ and $ B $, the balance point is achieved at 64 cm. When connected between $ A $ and $ C $, the balance point occurs at 8 cm. Now, we need to determine the balance point when the potentiometer is connected between $ B $ and $ C $.

Calculation

Initial Conditions:

When connected between $ A $ and $ B $:

$ V_1 = k \times l_1 = 64k $

When connected between $ A $ and $ C $:

$ V_1 - V_2 = k \times l_2 = 8k $

Solve for $ V_2 $:

Rearrange the equation:

$ 64k - V_2 = 8k $

Solve for $ V_2 $:

$ V_2 = 64k - 8k = 56k $

Determine the balance point when connected between $ B $ and $ C $:

We know $ V_2 = k \times l_2 $, where $ l_2 $ is the distance we need to find.

Therefore:

$ k \times l_2 = 56k $

Solve for $ l_2 $:

$ l_2 = 56 \text{ cm} $

Thus, the balance point when the potentiometer is connected between $ B $ and $ C $ will be 56 cm.

The direction of current is from $A$ to $B$ which means point $A$ is at higher potential.

For section $A B$,

$ \begin{aligned} & V=I R=1 \times 1.5 \\ & V=+1.5 \mathrm{~V} \text { at } A \end{aligned} $

Now for section $B C$

$ \begin{gathered} V=I R \quad \text { (at } C \text { let potential beX) } \\ (2 V+X)=I R \Rightarrow I=\frac{2 V+X}{2.5 \Omega} \end{gathered} $

We know from diagram $I=1 A$

For $I$ to be 1 the $V / R$ ratio should be $L$.

$ \begin{array}{ll} \because & 2.5=2+X \Rightarrow X=2-2.5 \\ \Rightarrow & X=-0.5 \mathrm{~V} \end{array} $

Since they have low temperature coefficient of resistance, their resistance remains almost constant.

$R = {R_1} + {R_2}$

$ \Rightarrow {{{l_1} + {l_2}} \over {\sigma A}} = {{{l_1}} \over {{\sigma _1}A}} + {{{l_2}} \over {{\sigma _2}A}}$

$ \Rightarrow {2 \over \sigma } = {1 \over {{\sigma _1}}} + {1 \over {{\sigma _2}}}$

$ \Rightarrow \sigma = {{2{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}$

Statement I : ${R_{1\,part}} = {{80} \over 4} = 20\,\Omega $

$ \Rightarrow {R_{eff}} = {{20} \over 4} = 5\,\Omega $

Statement II : Ratio $ = {{{{{{(\Delta V)}^2}} \over {3R}}} \over {{{{{(\Delta V)}^2}} \over {2R}}}}$

$ = {2 \over 3}$

We know, Resistivity $\rho=\frac{m}{n e^2 \tau}$

Where $\tau$ is relaxation time As temperature $\uparrow, \tau \downarrow, \rho \uparrow, \mathrm{R} \uparrow, \mathrm{i} \downarrow, \mathrm{v}_{\mathrm{d}} \downarrow$ as $i=$ ne $\mathrm{A} v_{\mathrm{d}}$

Statement (A) is correct and hence statement (E) is incorrect.

$ \mathrm{R}=\rho \frac{l}{A}, I=\frac{V}{R}=\frac{V A}{\rho l}=n e A v_d \rightarrow v_d=\text { constant } $

For a given $\mathrm{V}, v_{\mathrm{d}}$ is independent of $\mathrm{A}$. Hence, statement (B) is incorrect

Form above $v_d \propto$ V statement (C) is also incorrect. Since $v_d \propto \frac{1}{\ell}$ as seen from above statement, so statement (D) is correct.

$\Delta V = 0 \Rightarrow {{2\varepsilon } \over {{r_1} + {r_2} + R}}{r_1} = \varepsilon $

$ \Rightarrow R = {r_1} - {r_2}$

Balance wheatstone

$ \Rightarrow {R_{eff}} = {{3 \times 6} \over {3 + 6}} \times 2 + 2$

$ = 6\,\Omega $

$ \Rightarrow I = {V \over R} = 1\,A$

The grouping of resistance is a wheatstone bridge

So, ${R_{net}} = 4\,\Omega $

So, $i = {V \over {{R_{net}}}} = 10\,A$

Effective $R = \left[ {5 + {{5 \times 10} \over {5 + 10}} + 10} \right]\,k\Omega $

$ = {{275} \over {15}}k\Omega $

$ \Rightarrow \Delta {V_{AB}} = 15\,mA \times {{275} \over {15}}k\Omega $

$ = 275$ V

Electric displacement $(\overrightarrow D ) = {\varepsilon _0}\overrightarrow E $

$ \Rightarrow [\overline D ] = [{\varepsilon _0}][\overline E ]$

$ = [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^1}{A^{ - 1}}{T^{ - 3}}]$

$[\overline D ] = [{L^{ - 2}}{T^1}{A^1}]$

[Surface charge density] $ = {{[Q]} \over {[A]}}$

$[\sigma ] = [AT{L^{ - 2}}]$

$ \Rightarrow \overrightarrow D $ and $[\sigma ]$ have same dimensions

Initially, copper wire radius (r₁) = 9 mm

Resistance (R) = 14 $\Omega$

We know, $R = {{\rho L} \over A} = {{\rho L} \over {\pi r_1^2}} = 14$

Now this copper wire is replaced by 7 parallel copper wire of resistance R₁.

$\therefore$ Equivalent resistance of 7 parallel copper wire,

${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}}$

$ \Rightarrow {1 \over {{R_{eq}}}} = {7 \over {{R_1}}}$

$ \Rightarrow {R_{eq}} = {{{R_1}} \over 7}$

$ = {1 \over 7} \times {{\rho L} \over {\pi r_2^2}}$

$ = {1 \over 7} \times {{\rho L} \over {\pi {{\left( {{{{r_1}} \over 3}} \right)}^2}}}$ [as ${r_2} = {{{r_1}} \over 3}$ ; ${r_1} = 9$ m and ${r_2} = 3$]

$ = {1 \over 7} \times {{\rho L} \over {\pi r_1^2}} \times 9$

$ = {1 \over 7} \times 14 \times 9$

$ = 18\,\Omega $

From diagram

${i_p} = {E \over {2 + {r \over 2}}}$ and ${i_s} = {{2E} \over {2 + 2r}}$

given ${i_p} = {i_s}$

${1 \over {2 + {r \over 2}}} = {1 \over {1 + r}}$

$1 + r = 2 + {r \over 2}$

$r = 2\,\Omega $

R₁₀ = 2 = R₀(1 + $\alpha$ $\times$ 10)

R₃₀ = 3 = R₀(1 + $\alpha$ $\times$ 30)

On solving

$\alpha$ = 0.033/$^\circ$C

From the given setup

$y \times {R_G} = (x - y)({R_S})$

$ \Rightarrow y \times 72 = (x - y) \times 8$

$ \Rightarrow 9y = x - y$

$ \Rightarrow y = {x \over {10}}$ or 10% of x

Initially 5 $\Omega$ and 5 $\Omega$ are in series and then in parallel with 10 $\Omega$ this pattern continues thus

R_net = 5 $\Omega$

$R = {{\rho l} \over A}$

Also volume will remain constant

i.e., Al = constant $ \Rightarrow A \propto {1 \over l}$

$\therefore$ $R \propto {l^2}$

${{\Delta R} \over R} = 2{{\Delta l} \over l} = 0.8$

$I = {{2\varepsilon } \over {R + {r_1} + {r_2}}}$

As per the question,

${{2\varepsilon } \over {R + {r_1} + {r_2}}} \times {r_2} - \varepsilon = 0$

$ \Rightarrow R = {r_2} - {r_1}$

According to the information, current through galvanometer = nK

$ \Rightarrow {S \over {S + G}}i = nK$

$ \Rightarrow i = {{nK(S + G)} \over S}$

The circuit for the given situation is:

Since G and S are in parallel,

$\Rightarrow$ ${i \over 3}$ $\times$ G = ${2i \over 3}$ $\times$ S

$\Rightarrow$ G = 2S

$\Rightarrow$ G equals twice the value of shunt resistance.

${R_{eq}} = {{2 \times 4} \over {2 + 6}} + 6 = {{22} \over 3}$

$\Rightarrow$ A and B are in parallel and C is in series.

${{1.5 \times 10} \over {10 + {r \over 2}}} = 1.2$

$\Rightarrow$ r = 5 $\Omega$