Current Electricity

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

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

Here, resistance of the galvanometer

$R_G=60\Omega$

When the galvanometer is shunted by a resistance r, its effective resistance

$R_P=\frac{R_G r}{R_G+r}=\frac{60 \times 0.02}{60+0.02} \approx 0.02 \Omega$

Total resistance of the circuit $=R+R_P=R+0.02$

$\begin{aligned} \text { Current, } I & =\frac{5}{R+0.02} \Rightarrow 1=\frac{5}{R+0.02} \\ R+0.02 & =5 \Rightarrow R=5-0.02=4.98 \approx 5 \Omega \end{aligned}$

$\because V_{A B}=\left(I-I_g\right) R=I_g \cdot G$

where, G = voltmeter (resistance).

$\begin{array}{ll} \text { Given, } & V_{A B}=120 \mathrm{~V}, I=2 \mathrm{~A}, R=75 \Omega \\\\ \Rightarrow & 120=\left(2-I_g\right) 75 \Rightarrow I_g=0.4 \mathrm{~A} \\\\ \text { Now, } & V_{A B}=I_g G \Rightarrow G=\frac{120}{0.4}=300 \Omega \end{array}$

During charging, $E=V+i r$ (due to reversed current). In case of charging emf of a cell is less than its terminal voltage while in case of discharging emf is greater than terminal voltage.

Switching results in high decay/growth rate of current which results in a high. Current when bulb is turned off (due to back emf). So, a bulb is must likely to get fused when it is just turned off.

$\text { Given, } J_B=\frac{J_A}{2}$

$\begin{aligned} &\begin{aligned} \text { i.e., } 2 \times I\left(\pi r_A^2\right) & =I\left(\pi r_B^2\right) \\ \Rightarrow \quad r_B & =\left(r_A\right) \sqrt{2}=0.5 \times 1.414 \\ & =0.7 \mathrm{~cm} \end{aligned} \end{aligned}$

Figure is equivalent to the one shown below.

It is a Wheatstone’s bridge in which

$\begin{aligned} & P=\frac{6 \times 3}{6+3}=2 \Omega \\\\ & Q=\frac{8}{2}=4 \Omega \\\\ & R=\frac{4}{2}=2 \Omega \text { and } S=\frac{20 \times 5}{20+5}=4 \Omega \end{aligned}$

We find that $\frac{P}{Q}=\frac{2}{4}=\frac{R}{S}$

i.e. the bridge is balanced. Effective resistance between $p$ and $q$

$\begin{aligned} R_{A B} & =\frac{(P+Q)(R+S)}{P+Q+R+S}=\frac{(2+4)(2+4)}{2+4+2+4} \\\\ & =\frac{6 \times 6}{12}=3 \Omega \end{aligned}$

It is clear that the two cells oppose each other hence the effective emf in closed circuit is $18-12=6 \mathrm{~V}$ and net resistance is $1+2=3 \Omega$ (because in the closed circuit the internal resistance of two cells are in series)

The current in circuit will be in direction of arrow shown in figure

$\begin{aligned} I & =\frac{\text { effective emf }}{\text { total resistance }} \\\\ & =6 / 3=2 \mathrm{~A} \end{aligned}$

The potential difference across V will be same as the terminal voltage of either cell

Since, current is drawn from the cell of $18 \mathrm{~V}$ hence

$\begin{aligned} V_1 & =E_1-I r_1=18-(2 \times 2) \\\\ & =18-4=14 \mathrm{~V} \end{aligned}$

When a voltmeter is connected across the two terminals of a cell, it draws a small current from the cell, so it measures terminal potential difference between the two terminals of the cell, which is always less than the emf of the cell on the other hand, when a potentiometer is used for the measurement of emf of cell, it does not draw any current from the cell. Hence, it accurately measures the emf of cell. Thus, a potentiometer is preferred over a voltmeter.