$v_{d} = \frac{eE}{m}\tau$
$i = \frac{12}{1+6+3} = 1.2 \text{ A}$
$V_{A} - V_{B} = 12 - (1.2 \times 1) = 10.8 \text{ V}$
$V_{D} - V_{R} = 1.2 \times 6 = 7.2 \text{ V}$
$V_{C} - V_{B} = \frac{10.8 \times 2}{3 \times 1} \text{ V} = 7.2 \text{ V}$
$V_{D} - V_{C} = 0$
Let potential at point O be smaller than each potential
$\frac{10 - V_{0}}{10} + \frac{5 - V_{0}}{30} + \frac{6 - V_{0}}{20} = 0$
On solving we get $V_{0} = 8 \text{ volt}$
$i_{1} = \frac{10 - 8}{10} = 0.2 \text{ A}$
For $i_{max}$, $20~\Omega$ is excluded.
So $R_{eq} = 10~\Omega$
$i = \frac{V}{R} = \frac{5}{10} = 0.5 \text{ A}$
When $K_{1}$ and $K_{2}$ are closed:
In parallel, current $\propto \frac{1}{\text{resistance}}$
If current in $3\Omega$ is $I_{1} = 6\text{A}$, the current in $6\Omega$ is $I_{2} = 3\text{A}$.
Total current $I = 6 + 3 = 9\text{A}$
Total resistance between A and B $= 8 + \frac{3 \times 6}{3 + 6} = 10~\Omega$
Let V be the applied voltage. Then, $V = I \cdot R_{t} = 9 \times 10 = 90\text{V}$
When key $K_{2}$ is opened, resistance in the circuit becomes $8 + 3 = 11~\Omega$
Current in the ammeter $= \frac{90}{11} \text{ A}$
From graph it is clear that slope of the line B is lower than the slope of line A.
Also slope = Resistance i.e. $\tan \theta = \frac{\Delta V}{\Delta I} = R$
$\theta_{1} < \theta_{2} \implies \tan \theta_{1} < \tan \theta_{2} \implies R_{B} < R_{A}$
$(3l)^{2} \times 5 = 10$
$(4l)^{2} \times 2 = P$
$\frac{9}{16} \times \frac{5}{2} = \frac{10}{P} \implies P = \frac{10 \times 16 \times 2}{9 \times 5} = 7.1 \text{ cal/s}$
In balancing condition:
$\frac{R_{1}}{R_{2}} = \frac{l}{100-l}$ or $\frac{R_{1}}{l} = \frac{R_{2}}{100-l}$
$\frac{X}{50} = \frac{Y}{50} \implies Y = X$
and $\frac{4X}{l} = \frac{Y}{100-l} \implies l = 80 \text{ cm}$
When the voltmeter is not connected, current $I = \frac{30}{20+60+40} = \frac{1}{4} \text{ A}$
Potential difference over $60\Omega = 60 \times \frac{1}{4} = 15 \text{ V}$
When voltmeter is connected across $60\Omega$, the net resistance in the circuit becomes $R^{\prime} = 20 + \frac{120 \times 60}{120+60} + 40 = 20 + 40 + 40 = 100~\Omega$
Current $I^{\prime} = \frac{30}{100} = \frac{3}{10} \text{ A}$
Potential difference across parallel combination of $60\Omega$ and $120\Omega = \frac{3}{10} \times 40 = 12 \text{ V}$
Percentage error $= \frac{15 - 12}{15} \times 100 = 20\%$
Thus, the reading is less by 20%.
Current density $J = ar^{2}$
Current $I = \int J~dA$
$I = \int_{R/3}^{R/2} ar^{2}(2\pi r dr)$ [where $dA = 2\pi r dr$]
$= 2\pi a \left[\frac{r^{4}}{4}\right]_{R/3}^{R/2} = \frac{65\pi aR^{4}}{2592}$
$R = \frac{220 \times 220}{500} = \frac{484}{5} = 96.8~\Omega$
Detailed solution not required.
(when switch is open)
$i = \frac{V}{3R} + \frac{V}{3R} = \frac{2V}{3R} = 0.6\frac{V}{R}$
(when switch is closed)
$i_{1} = \frac{V}{\frac{2R}{3} + \frac{2R}{3}} = \frac{3V}{4R} = 0.75\frac{V}{R}$
So, reading will increase.
All the branches are in parallel when $B_{2}$ gets fused, total resistance will increase and current will decrease.
$4R = \rho\frac{(l-x)}{A} + \rho\frac{(0.5l+x)}{A^{\prime}}$
$4\rho\frac{l}{A} = \rho\frac{l-x}{A} + \rho\frac{(0.5l+x)^{2}}{xA}$
$\frac{x}{l} = \frac{1}{8}$
$I = \frac{Q}{t} = qf$
$= 1.6 \times 10^{-19} \times 6.6 \times 10^{15} = 10.5 \times 10^{-4} \text{ A} = 1\text{ mA}$
Resistivity of a wire depends on material & temperature.
$I = \frac{ne}{t}$
$= 2 \times 3.2 \times 10^{18} \times 1.6 \times 10^{-19} + 3.6 \times 10^{18} \times 1.6 \times 10^{-19}$
$= 16 \times 10^{-1} \text{ Amp}$
$V_{d} = \frac{i}{neA}$
$= \frac{20}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$= 1.25 \times 10^{-3} \text{ m/sec}$
$q = it$
$q = 5 \times 60 \text{ min} = 300\text{ C}$
Relaxation time ($\tau$)
$\tau = \frac{\lambda}{V_{rms}}$
$V_{rms} \propto \sqrt{\text{Temp}}$
Hence $\tau \propto \frac{1}{\sqrt{T}}$
For maximum current:
$R = r_{net}$
$3 = \frac{mr}{n} \implies 3 = \frac{m}{n} \times 0.5$
$\frac{m}{n} = 6 \implies m = 6n$
$mn = 24$ (Given)
$6n^{2} = 24 \implies n = 2$, $m = 12$
$R_{series} = R_{1} + R_{2} + R_{3}$
If R is cut into ten equal parts then resistance of 1 part is $\frac{R}{10}$.
The equivalent resistance for the overall combined structure becomes $R^{\prime}_{eq} = \frac{R}{5 \times 5} = \frac{R}{25}$
Current in the circuit
$i = \frac{8}{6} = \frac{4}{3} \text{ Amp}$
$V_{E} + 1 \times \frac{4}{3} = V_{C}$
$V_{E} = -\frac{4}{3} \text{ V}$
$R_{1} = 4~\Omega$
$R_{2} = 2~\Omega$
$V_{2} = \frac{R_{2}}{R_{1} + R_{2}}V = \frac{2}{2+4} \times 3 \implies V_{2} = 1 \text{ V}$
$V_{AB} = 4 = \frac{5X + 2 \times 10}{x + 10}$
$x = 20~\Omega$
In balanced Wheatstone bridge BD arm is ineffective hence no current in BD arm.
$\sum V = 0$
Conservation of energy
$t = \frac{t_{1}t_{2}}{t_{1} + t_{2}} = \frac{10 \times 20}{10 + 20} = \frac{20}{3} \text{ min}.$
$x \times 60 = 40 \times 6 \implies x = 4~\Omega$
$i = \frac{5}{10} + \frac{5}{10} = 1\text{ amp}$
$\mu = \frac{5 \times 10^{-6}}{10} \times 14 = 2 \times 10^{-7}$
Detailed solution not required.
Resistance will be same.
$\frac{\rho_{1}}{A_{1}} = \frac{\rho_{2}}{A_{2}} \implies \frac{\rho_{1}}{\rho_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}}$
Detailed solution not required.
$R_{4} = 300~\Omega$
$I_{1} = \frac{2 \times 1000}{75} = 26.7 \text{ mA}$
$I_{2} = \frac{2}{500} \times 1000 = 4 \text{ mA}$
$V_{B} - 4 + 6 + 2 = V_{A}$
$V_{B} - V_{A} = -4\text{ V}$
$V_{C} - 1 + 1 - 1 - 2 - 1 + 1 - 3 = V_{D}$
$V_{C} - V_{D} = 6\text{ V}$
$R_{x} = \rho\frac{L}{\pi r^{2}} = \rho\frac{V}{\pi^{2} r^{4}}$
$R_{y} = \rho\frac{V}{(r/3)^{4}} = 81 R_{x}$
$R_{eq} = 14~\Omega$
$i_{\text{through } 5\Omega} = \frac{28}{14} = 2 \text{ amp}$
$V_{A} - 0.5 \times 10 - 1 \times 2 = V_{B}$
$V_{A} - V_{B} = 7\text{ V}$
Detailed solution not required.