Current Electricity Practice Questions

$R_1=R_0 [1+\alpha _1 \Delta t]$ $R_2=R_0 [1+\alpha _2 \Delta t]$ $R=R_1+R_2$ $=R_0 [2+ (\alpha _1+\alpha _2) \Delta t]$ $=2 R_0 [1+ (\;\frac{\alpha _1+\alpha _2}{2}) \Delta t]$ $\alpha _{e q}=\;\frac{\alpha _1+\alpha _2}{2}$ In Parallel, $\;\frac{1}{R}=\;\frac{1}{R_1}+\;\frac{1}{R_2}$ $=\;\frac{1}{R_0 [1+\alpha _1 \Delta t] }+\;\frac{1}{R_0 [1+\alpha _2 \Delta t] }$ $\Rightarrow \;\frac{1}{\frac{R_0}{2} (1+\alpha _{e q} \Delta t) }=\;\frac{1}{R_0 (1+\alpha _1 \Delta t) }+\;\frac{1}{R_0 (1+\alpha _2 \Delta t) }$ $2 (1-a_{e q} \Delta t) = (1-\alpha _1 \Delta t) (1-\alpha _2 \Delta t)$ $\therefore \alpha _{e q}=\;\frac{\alpha _1+\alpha _2}{2}$

As shown above $E=\;\frac{\rho I}{2 \pi r^2}$

Applying Kirchhoff's loop law in $A B P_2 P_1 A$ we get$-2 i+5-10 i_1=0 \;\; . . .(i)$Again applying kirchoffs loop law in $P_2 C D P_1 P_2$ we get,$10 i_1+2-i+i_1=0$From $(i)$ and $(i i) 11 i_1+2- [\;\frac{5-10 i_1}{2}] =0$ $\Rightarrow i_1=\;\frac{1}{32}$ A from $P_2$ to $P_1$

According to the condition of balancing$\;\frac{55}{20}=\;\frac{R}{80} \Rightarrow R=220 Ω$

Let $j$ be the current density. Then $j \times 2 \pi r^2=I \Rightarrow j=\;\frac{I}{2 \pi r^2}$ $\therefore E=\rho j=\;\frac{\rho I}{2 \pi r^2}$ Now, $\Delta V_{B C}^{'}=-∫_{a+b}^{a} \vec{E} \cdot \vec{d r}=-∫_{a+b}^{a} \;\frac{\rho I}{2 \pi r^2} d r$ $=-\;\frac{\rho I}{2 \pi } [-\;\frac{1}{r}] _{a+b}^a$ $=\;\frac{\rho I}{2 \pi a}-\;\frac{\rho I}{2 \pi (a+b)}$ On applying superposition as mentioned we get $\Delta V_{B C}=2 \times \Delta V_{B C}=\;\frac{\rho I}{\pi a}-\;\frac{\rho I}{\pi (a+b)}$

We know that $R_t=R_0(1+\alpha t)$ where $R_t$ is the resistance of the wire at $t^{\circ} C$, $R_0$ is the resistance of the wire at $0^{\circ} C$ and $\alpha$ is the temperature coefficient of resistance $\Rightarrow R_{50}=R_0(1+50 \alpha )$ ......(i) $R_{100}=R_0(1+100 \alpha )$ ......(ii) From $(i), R_{50}-R_0=50 \alpha R_0$......(iii) From $(i i), R_{100}-R_0=100 \alpha R_0$......(iv) Dividing $(i i i)$ by $(i v)$, we get $\;\frac{R_{50}-R_0}{R_{100}-R_0}=\;\frac{1}{2}$ Here, $R_{50}=5 Ω$ and $R_{100}=6 Ω$ $\therefore \;\frac{5-R_0}{6-R_0}=\;\frac{1}{2}$ or, $6-R_0=10-2 R_0$ or, $R_0=4 Ω$.

$\;\frac{P}{Q}=\;\frac{R}{S}$ where $S=\;\frac{S_1 S_2}{S_1+S_2}$

The network of resistors is a balanced wheatstone bridge. The equivalent circuit is${\text{R}}_{e q}=\;\frac{15 \times 30}{15+30}=10 Ω$ $\Rightarrow I=\;\frac{V}{R}=\;\frac{5}{10}=0.5 {\text{A}}$

The resistance of the bulb is $R=\;\frac{V^2}{P}=\;\frac{(220)^2}{100}$The power consumed when operated at $110 {\text{V}}$ is$P=\;\frac{(110)^2}{(220)^2 {/} 100}=\;\frac{100}{4}=25 {\text{W}}$

At cold junction, current flows from Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series).

$R_1=R_0[1+\alpha \times 100]=100$.....(1)$R_2=R_0[1+\alpha \times T]=200$.....(2) On dividing we get $\;\frac{200}{100}=\;\frac{1+\alpha T}{1+100 \alpha }$ $\Rightarrow 2=\;\frac{1+0.005 T}{1+100 \times 0.005}$ $\Rightarrow T=400^{\circ} C$ NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $R=R_0 e^{\alpha \Delta T}$

Mass deposited $m=Z q \Rightarrow Z \propto \;\frac{1}{q} \Rightarrow \;\frac{Z_1}{Z_2}=\;\frac{q_2}{q_1}$.....(i) Also $q=q_1+q_2$....(ii) $\Rightarrow \;\frac{q}{q_2}=\;\frac{q_1}{q_2}+1$ (Dividing (ii) by $q_2$) $\Rightarrow q_2=\;\frac{q}{1+\;\frac{q_1}{q_2}} \;\;$......(iii) From equations $(i)$ and $(i i i), q_2=\;\frac{q}{1+\;\frac{Z_2}{z_1}}$

KEY CONCEPT : Resistance of Galvanometer,$G=\;\frac{\;\text{ Current sensitivity }\;}{\;\text{ Voltage sensityvity }\;} \Rightarrow G=\;\frac{10}{2}=5 Ω$Here $i_g=$ Full scale deflection current $=\;\frac{150}{10}=15 {\text{mA}}$ $V=$ voltage to be measured $=150$ volts(such that each division reads 1 volt)$\Rightarrow R=\;\frac{150}{15 \times 10^{-3}}-5=9995 Ω$

$I=\;\frac{E}{R+r}$, Internal resistance $(r)$ iszero, $I=\;\frac{E}{R}=$ constant.

$i R=2=12-500 i$ $\therefore i=\;\frac{1}{50}$ $\therefore \;\frac{1}{50} \times R=2$ $R=100 Ω$

${\text{I}}=\;\frac{2 \epsilon }{R+R_1+R_2}$Potential difference across second cell$\;=V=\epsilon - {\text{I}} R_2=0$ $\;\epsilon -\;\frac{2 \epsilon }{R+R_1+R_2} \cdot R_2=0$ $\;R+R_1+R_2-2 R_2=0$ $\;R+R_1-R_2=0$ $\; \therefore R=R_2-R_1$

The internal resistance of the cell,$\;r= (\;\frac{ℓ_1-ℓ_2}{ℓ_2}) \times R$ $\;=\;\frac{240-120}{120} \times 2=2 Ω$

$H=\;\frac{V^2 t}{R}$Resistance of half the coil $=\;\frac{R}{2}$ $\therefore$ As $R$ reduces to half, ${ }^{'} H^{'}$ will be doubled.

$P=V i=\;\frac{V_2}{R}$ $R_{\;\text{hot }\;}=\;\frac{V^2}{P}=\;\frac{200 \times 200}{100}=400 Ω$ $R_{\;\text{cold }\;}=\;\frac{400}{10}=40 Ω$

Thermistors are usually made of metal-oxides with high temperature coefficient of resistivity.