Current Electricity

Sol. Circuit can be draw as,

⇒ Terminal voltage of battery $V=E$ - ir

$ =12-0.6 \times 2=10.8 \mathrm{~V} $

→ Each side will have resistance of $1 \Omega$

→ It is balanced wheat stone bridge

→ No current in the resistance of $2 \Omega$

$\rightarrow \quad\left[R_{\text {effective }}\right]_{A C}=1 \Omega$

$ \begin{aligned} & \rightarrow \quad I=\frac{E}{R_{\text {eff. }}}=\frac{2}{1} \\ & \rightarrow \quad I=2 \mathrm{~A} \end{aligned} $

Given:

• Rated power $P_1 = 400\,\text{W}$

• Rated voltage $V_1 = 220\,\text{V}$

• New voltage $V_2 = 200\,\text{V}$

Using the relation

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

First, find the resistance of the heater:

$ R = \frac{V_1^2}{P_1} = \frac{220^2}{400} = \frac{48400}{400} = 121\,\Omega $

Now find the new power at $200\,\text{V}$:

$ P_2 = \frac{V_2^2}{R} = \frac{200^2}{121} = \frac{40000}{121} \approx 330.6\,\text{W} $

So, approximately,

$ P_2 \approx 331\,\text{W} $

Therefore, the correct answer is:

$ \boxed{\text{Option C: } 331\,\text{W}} $

$ \begin{aligned} i_G & =1 \mathrm{~mA}=0.001 \mathrm{~A} \\ i_S & =10-i_G \simeq 10 \mathrm{~A} \end{aligned} $

Both shunt resistance and galvanometer are in parallel connection

$ \begin{array}{ll} \therefore & i_S r_S=i_G R_G \\ \Rightarrow & 10 \times r_s=0.001 \times 100 \\ \Rightarrow & r_s=0.01 \Omega \end{array} $

• Position of null point will not change when galvanometer ( $G$ ) and the cell ( $E$ ) are interchanged.

• There will be no deflection in galvanometer only at balance point.

In an unbalanced meter bridge, if $E$ and $G$ are interchanged mutually, then the deflection in galvanometer may be towards left-side or right-side.

$ \begin{aligned} \\& R_{e q}=R_0[1-\alpha \Delta T]+R_0[1+\alpha \Delta T] & \\ = & 2 R_0=\text { constant } \\ \therefore & I_A=\frac{V}{2 R_0} \rightarrow \text { constant } \end{aligned} $

$ \begin{aligned} & R_{\mathrm{eq}}=\frac{R_0(1+\alpha \Delta T) R_0(1-\alpha \Delta T)}{R_0(1+\alpha \Delta T)+R_0(1-\alpha \Delta T)} \\ & R_{\mathrm{eq}}=\frac{R_0\left(1-\alpha^2(\Delta T)^2\right)}{2 R_0} \end{aligned} $

$R_{\text {eq }}$ will decrease with rise in temperature

$\therefore I_B=\frac{V_0}{R_{\mathrm{eq}}}$ increases with rise in temperature

$\therefore$ its equivalent $R^{\prime}=\frac{4 \times 8}{12}=\frac{8}{3} \Omega$

Circuit can be redrawn as

$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{8}{3}+\frac{1}{3}+1.5+5.5 \\ & =10 \Omega \\ & i=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{5}{10}=0.5 \mathrm{~A} \end{aligned}$

After cutting the wire into 8 equal pieces:

Resistance of each piece:

Each piece has a resistance $R' = \frac{R}{8}$.

Parallel Combination in Each Set:

Each set consists of 4 pieces connected in parallel:

Resistance of each set:

$ R'' = \frac{R'}{4} = \frac{R}{32} $

Connecting the Sets in Series:

With both sets connected in series, the equivalent resistance is calculated as:

$ R_{\text{eq}} = R'' + R'' = 2 \times \frac{R}{32} = \frac{R}{16} $

Thus, the net effective resistance of the combination is $\frac{R}{16}$.

$\begin{aligned} & R_{A B}=(1 \Omega / 3 \Omega) \text { in series with }(2 \Omega / 4 \Omega) \\ & =\frac{3 \times 1}{3+1}+\frac{2 \times 4}{2+4} \\ & =\frac{3}{4}+\frac{8}{6}=\frac{9+16}{12}=\frac{25}{12} \Omega \end{aligned}$

Now total current through cell

$\begin{aligned} & I=\frac{50}{25 / 12}=24 \mathrm{~A} \\ & I_{1 \Omega}=\frac{3}{4} \times 24=18 \mathrm{~A}, I_{3 \Omega}=\frac{1}{4} \times 24=6 \mathrm{~A} \\ & I_{2 \Omega}=\frac{4}{6} \times 24=16 \mathrm{~A}, I_{4 \Omega}=\frac{2}{6} \times 24=8 \mathrm{~A} \end{aligned}$

Using junction rule at $C, I_{C D}=18-16=2 \mathrm{~A}$ (From $C$ to $D$)

To solve this problem, we need to understand the relationship between the current, the drift velocity, and the cross-sectional area of the wire. The electric current $I$ in a wire is given by

$I = n e A v_d$

where:

• $n$ is the number density of electrons,
• $e$ is the charge of an electron,
• $A$ is the cross-sectional area of the wire, and
• $v_d$ is the mean drift velocity of the electrons.

Let's denote the current in the thicker wire as $I_1 = 100 \mathrm{~mA} = 0.1 \mathrm{~A}$ and the current in the thinner wire as $I_2 = 200 \mathrm{~mA} = 0.2 \mathrm{~A}$.

The diameter of the thicker wire is $d$, so its cross-sectional area, $A_1$, is

$A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$

For the thinner wire, the diameter is $\frac{d}{2}$, so its cross-sectional area, $A_2$, is

$A_2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16}$

Using the relation for current in the thicker wire, we have

$I_1 = n e A_1 v = n e \left(\frac{\pi d^2}{4}\right) v = 0.1 \mathrm{~A}$

Let the mean drift velocity in the thinner wire be denoted by $v'$. For the thinner wire:

$I_2 = n e A_2 v' = n e \left(\frac{\pi d^2}{16}\right) v' = 0.2 \mathrm{~A}$

Now, we can equate the expressions for current and solve for $v'$:

$\frac{\pi d^2}{16} n e v' = 0.2 \mathrm{~A}$

$v' = \frac{0.2 \mathrm{~A}}{n e \left(\frac{\pi d^2}{16}\right)}$

From the equation for the thicker wire:

$0.1 \mathrm{~A} = \frac{\pi d^2}{4} n e v$

$v = \frac{0.1 \mathrm{~A}}{n e \left(\frac{\pi d^2}{4}\right)}$

Dividing the two equations, we get

$\frac{v'}{v} = \frac{0.2 \mathrm{~A}}{\left(\frac{\pi d^2}{16}\right) n e} \times \frac{(\frac{\pi d^2}{4}) n e}{0.1 \mathrm{~A}} = \frac{0.2 \mathrm{~A} \times 4}{0.1 \mathrm{~A} \times 16} = \frac{4}{1} = 8$

Thus,

$v' = 8 v$

The mean drift velocity of electrons in the thinner wire is 8 times the mean drift velocity in the thicker wire. Therefore, the correct option is:

Option B: $8 v$

$R_0=10 \Omega$

After stretching its length upto $2l$

$\begin{aligned} R_1 & =n^2 R_0 \\ & =4 R_0 \\ & =40 \Omega \end{aligned}$

$R_{A B}=\frac{20 \times 20}{20+20}=10 \Omega$

$\begin{aligned} & \frac{8}{x}=\frac{12}{40-x} \\ & \frac{2}{x}=\frac{3}{40-x} \\ & 80-2 x=3 x \\ & 16=x \end{aligned}$

from $B$

$=40-16=24 \mathrm{~cm}$

$ \begin{aligned} \text { Current in circuit } i & =\frac{10}{4+1}=2 \mathrm{~A} \\ \text { Terminal voltage } & =E-i R \\ & =10-2 \times 1=8 \mathrm{~V} \end{aligned} $

To solve this problem, we first need to understand how the resistance changes when we cut the wire into equal parts and how it behaves when connected in different configurations (series and parallel).

Given:

• Original length of the wire, $ l $.
• Total resistance of the wire $ R = 100 \Omega $.
• The wire is divided into 10 equal parts.

Resistance of each part:

Since the wire is divided into 10 equal parts, the length of each part is $\frac{l}{10}$. Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as $ r $, is $\frac{1}{10}$th of the total resistance:

$ r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega $

First 5 parts in series:

When resistors are connected in series, the total resistance is the sum of the individual resistances:

$ R_{\text{series}} = 5 \times 10 \Omega = 50 \Omega $

Next 5 parts in parallel:

When resistors are connected in parallel, the total resistance $ R_{\text{parallel}} $ can be calculated using the reciprocal formula:

$ \frac{1}{R_{\text{parallel}}} = \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} = 5 \times \frac{1}{10 \Omega} = \frac{5}{10 \Omega} = \frac{1}{2 \Omega} $

Thus,

$ R_{\text{parallel}} = 2 \Omega $

Final combination in series:

The total resistance of the combination, where the series and parallel groups are again connected in series, will be:

$ R_{\text{total}} = R_{\text{series}} + R_{\text{parallel}} = 50 \Omega + 2 \Omega = 52 \Omega $

Thus, the correct answer to the resistance of the final combination is:

Option B $52 \Omega$

To find the ratio of power outputs when two heaters with different power ratings are connected first in series and then in parallel, we need to understand how the total power output varies based on the type of connection.

Heater Specifications:

• Power of heater A $ (P_A) = 1 \, \text{kW} = 1000 \, \text{W} $
• Power of heater B $ (P_B) = 2 \, \text{kW} = 2000 \, \text{W} $

Scenario 1: Series Connection

When resistors (or heaters in this case) are connected in series, the total resistance ($R_{\text{series}}$) is the sum of the individual resistances ($R_A$ and $R_B$).

Using the formula for electrical power: $ P = \frac{V^2}{R} $,

where $ P $ is power, $ V $ is voltage, and $ R $ is resistance, we can express the resistance of each heater as:

$ R_A = \frac{V^2}{P_A} $

$ R_B = \frac{V^2}{P_B} $

Substitute the given power values:

$ R_A = \frac{V^2}{1000} $

$ R_B = \frac{V^2}{2000} $

Then the total resistance for the series connection is:

$ R_{\text{series}} = R_A + R_B = \frac{V^2}{1000} + \frac{V^2}{2000} = \frac{3V^2}{2000} $

The total power output in series ($P_{\text{series}}$) is:

$ P_{\text{series}} = \frac{V^2}{R_{\text{series}}} = \frac{V^2}{\frac{3V^2}{2000}} = \frac{2000}{3} \text{W} $

Scenario 2: Parallel Connection

For parallel connections, the total resistance ($R_{\text{parallel}}$) is given by:

$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{\frac{V^2}{1000}} + \frac{1}{\frac{V^2}{2000}} = \frac{3}{2V^2} $

Reformulate to find $R_{\text{parallel}}$:

$ R_{\text{parallel}} = \frac{2V^2}{3} $

And the total power output in parallel ($P_{\text{parallel}}$) is:

$ P_{\text{parallel}} = \frac{V^2}{R_{\text{parallel}}} = \frac{V^2}{\frac{2V^2}{3}} = \frac{3V^2}{2} $

However, simplifying,

$ P_{\text{parallel}} = \frac{3}{2} V^2 $

The ratio of powers is then:

$ \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{2000}{3}}{\frac{3V^2}{2}} $

Solving and simplifying,

$ \text{Ratio} = \frac{\frac{2000}{3}}{\frac{3 \times V^2}{2}} = \frac{2000 \times 2}{3 \times 3 \times V^2} = \frac{4000}{9V^2} $

Given that we know one ratio of the actual power values, we simplify further. For calculating power in simple terms, consider voltage to be normalized (taken out of the fraction):

$ \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{2000}{3}}{\frac{6000}{2}} = \frac{2000 \times 2}{3 \times 6000} = \frac{4000}{18000} = \frac{2}{9} $

Therefore, the ratio of the power outputs when the heaters are connected first in series and then in parallel is 2:9, which corresponds to Option B.

In option (1),

$ \frac{10}{15}=\frac{10}{5+R_D} $

The diode can conduct and have resistance $R_D=10 \Omega$ because diode have dynamic resistance. In that case bridge will be balanced.

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

If diameter becomes thrice then cross section area will become 9 times so

$\mathrm{R\propto \frac{1}{A}}$

Resistance will become $\frac{1}{9}$ times

$\mathrm{R'=\frac{81\Omega}{9}=9\Omega}$

$ \begin{aligned} & \mathrm{I}=\mathrm{neAV_{d }} \\ & \mathrm{V}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}=\frac{10}{10^{22} \times 1.6 \times 10^{-19} \times \pi \times 10^{-6}} \\ & \mathrm{~V}_{\mathrm{d}}=\frac{6.25}{\pi} \times 10^3 \mathrm{~m} / \mathrm{sec} \end{aligned} $

$\begin{aligned} & \mathrm{r}=\left(\frac{\ell_{\mathrm{o}}-\ell_{\mathrm{c}}}{\ell_{\mathrm{c}}}\right) \mathrm{R} \\ & 1=\left(\frac{330-\ell_{\mathrm{c}}}{\ell_{\mathrm{c}}}\right) \times 2 \\ & 3 \ell_{\mathrm{c}}=660 \\ & \ell_{\mathrm{c}}=220 \mathrm{~cm}\end{aligned}$

$\mathrm{i}=\frac{10-5}{10}=\frac{5}{10} \mathrm{~A}$

$=0.5 \mathrm{~A}$

from $A$ to $B$ through $E$.

For no reading galvanometer. Potential across it is same.

$\mathrm{i}_{400 \Omega} \Rightarrow \frac{10-2}{400}=\frac{8}{400}=\frac{1}{50}=\mathrm{i}_{\mathrm{R}}$

$\mathrm{i}_{\mathrm{R}} \Rightarrow \frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \Rightarrow \frac{2}{\mathrm{R}}=\frac{1}{50} \Rightarrow \mathrm{R}=100 \Omega$

$R=\left[22 \times 10^{3} \pm 5 \%\right] \Omega$

Acc. to color code

Third Band $\to$ Orange

(color code for digit 3 is orange)

$\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right]$

$6.8=2[1+\alpha(80-\alpha)]$

$\alpha=\frac{2.4}{80}=0.03 /{ }^{\circ} \mathrm{C}=3 \times 10^{-2} /{ }^{\circ} \mathrm{C}$

$\mathrm{I}_{\mathrm{S}}=\frac{E}{10 R}$ ..... (1)

$\mathrm{I}_{\mathrm{P}}=\frac{\mathrm{E}}{\mathrm{R} / 10}=\frac{10 \mathrm{E}}{\mathrm{R}}$ .... (2)

$\mathrm{n}=\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{S}}}=100 \Rightarrow \mathrm{n}=100$

$R = {1 \over G}$

Thus reciprocal of resistance (R) is conductance (G)

From Kirchhoff's loop law :

$V - Ir - IR = 0$

$ \Rightarrow I = \left( {{V \over {r + R}}} \right)$

Terminal potential difference across cell,

${V_{AB}} = IR$

${V_{AB}} = {{RV} \over {r + R}}$

$ = {{7.5 \times 4} \over {0.5 + 7.5}} = 3.75\,V$

Let net resistance of the given infinite network be 'R'

Now, the circuit can be modified as

Now, ${R_{net}} = R = 1 + 1 + {R \over {R + 1}}$

$\therefore$ $R = 2 + {R \over {R + 1}}$

${R^2} + R = 2R + 2 + R$

${R^2} - 2R - 2 = 0$

$R = \left( {{{2\, \pm \,\sqrt {4 + 8} } \over 2}} \right)$

$R = \left( {1 + \sqrt 3 } \right)\Omega $

Equivalent resistance across point AC

${R_{AC}} = {{{{{R_0}} \over 4} \times R} \over {{{{R_0}} \over 4} + R}} = {{R{R_0}} \over {{R_0} + 4R}}$

From voltage divider rule

${V_{AC}} = {{{R_{AC}}} \over {{R_{AC}} + {R_{CB}}}}{V_0} = {{{{R{R_0}} \over {{R_0} + 4R}}{V_0}} \over {{{R{R_0}} \over {{R_0} + 4R}} + {{3{R_0}} \over 4}}}$

$ = {{4R{R_0}{V_0}} \over {4R{R_0} + 3R_0^2 + 12R{R_0}}} = {{4R{V_0}} \over {3{R_0} + 16R}}$

For parallel combination

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

${{{P_1}} \over {{P_2}}} = {{{R_2}} \over {{R_1}}}$

$ \Rightarrow {{{P_1}} \over {{P_2}}} = {{200} \over {100}} = {2 \over 1}$

Resistance, $R = \rho {L \over A} = {L \over {\sigma A}}$

$ \Rightarrow \sigma = {L \over {RA}}$

Also current density $j = \sigma E = {{LE} \over {RA}}$

$J = {{10 \times 10} \over {10 \times \pi {{\left( {{{{{10}^{ - 2}}} \over {\sqrt \pi }}} \right)}^2}}} = {{100} \over {10 \times \pi \times \left( {{{{{10}^{ - 4}}} \over \pi }} \right)}}$

$ = {10^5}$ A/m²

Given, for a wire, $\frac{R}{l}=\frac{1}{2}$

Length of wire, $l=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$

$\therefore \quad R=\frac{l}{2}=2.5 \times 10^{-2} \Omega$

Potential difference, $V=1 \mathrm{~V}$ or $I R=1$

$I=\frac{1}{R}=\frac{1}{2.5 \times 10^{-2}}=\frac{100}{2.5}=40 \mathrm{~A}$

Given, current, $I=10 \mathrm{~A}$

Area of cross-section, $A=4 \times 10^{-6} \mathrm{~m}^2$

Density of conductors, $\rho=2.7 \mathrm{~gm} / \mathrm{cc}$

$=2.7 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$

Molecular weight of aluminium,

$M_w=27 \mathrm{~gm}=27 \times 10^{-3} \mathrm{~kg}$

If $n$ be the total number of electrons in the conductor per unit volume, then

$\begin{aligned} n & =\frac{\text { Total number of electrons }}{\text { Volume of conductor }(V)} \\ & =\frac{\text { Number of atoms per mole } \times \text { Number of moles }}{V} \\ & =\frac{\text { Avogadro Number }}{V} \times\left(\frac{M}{M_w}\right) \\ & =6 \times 10^{23} \times \frac{\rho}{M_w}=6 \times 10^{23} \times \frac{2.7 \times 10^3}{27 \times 10^{-3}} \\ \therefore \quad & n=6 \times 10^{28} \quad \text{.... (i)} \end{aligned}$

We know that, drift velocity

$\begin{aligned} v_d=\frac{I}{n e A} & =\frac{10}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-6}} \\ & =2.6 \times 10^{-4} \mathrm{~m} / \mathrm{s} \end{aligned}$

Given, $R_1=35 \Omega, l_2=2 l_1$

On increasing the length,

$\begin{aligned} m_1 & =m_2 \\ \therefore \quad \rho A_1 l_1 & =\rho A_2 l_2 \\ \pi r_1^2 l_1 & =\pi r_2^2 l_2 \\ \frac{r_1^2}{r_2^2} & =\frac{l_2}{l_1} \\ \frac{r_1^2}{r_2^2} & =\frac{2 l_1}{l_1} \\ \frac{r_1^2}{r_2^2} & =2 \quad \text{..... (i)}\\ \frac{R_1}{R_2} & =\frac{\rho \cdot \frac{l_1}{\pi r_1^2}}{\rho \cdot \frac{l_2}{\pi r_2^2}} \\ & =\frac{l_1}{l_2} \cdot \frac{r_2^2}{r_1^2}=\frac{l_1}{2 l_1} \cdot \frac{1}{2} \\ \frac{R_1}{R_2} & =\frac{1}{4} \\ & R_2=4 R_1=4 \times 35=140 \Omega \end{aligned}$

In the steady state, no current is passing through capacitor. Let the charge on each capacitor be q. Since, the current through galvanometer is zero.

$\therefore \quad I_1=I_2$

The potential difference between ends of galvanometer will be zero.

$\begin{array}{ll} \therefore & V_A-V_B=V_A-V_D \\ \therefore & I_1 R_1=\frac{q}{C_1} \quad \text{..... (i)} \end{array}$

$\begin{aligned} \text{Similarly,} \quad V_B-V_C & =V_D-V_C \\ I_2 R_2 & =\frac{q}{C_2} \quad \text{..... (ii)} \end{aligned}$

On dividing Eq. (i) by Eq. (ii), we get

$\begin{aligned} \frac{I_1 R_1}{I_2 R_2} & =\frac{q / C_1}{q / C_2}=\frac{C_2}{C_1} \\ \therefore \quad \quad \quad \frac{C_1}{C_2} & =\frac{R_2}{R_1} \end{aligned}$