Current Electricity Practice Questions

hence $R_{e q}=3 / 2$ $\therefore I=\;\frac{6}{3 / 2}=4 A$

$\rho _B=2 \rho _A$ $d_B=2 d_A$ $R_B=R_A \Rightarrow \;\frac{\rho _B ℓ_B}{A_B}=\;\frac{P_A ℓ_A}{A_A}$ $\therefore \;\frac{ℓ_B}{ℓ_A}=\;\frac{\rho _A}{\rho _B} \times \;\frac{d_B^2}{d_A^2}=\;\frac{\rho _A}{2 \rho _A} \times \;\frac{4 d_d^2}{d_A^2}=2$

In the first case $\;\frac{X}{Y}=\;\frac{20}{80}=\;\frac{1}{4}$In the second case $\;\frac{4 X}{Y}=\;\frac{ℓ}{100-ℓ} \Rightarrow ℓ=50$

Neutral temperature is the temperature of a hot junction at which $E$ is maximum.$\Rightarrow \;\frac{d E}{d \theta }=0$or $a+2 b \theta =0 \Rightarrow \theta =\;\frac{-a}{2 b}=-350$Neutral temperature can never be negative hence no $\theta$ is possible.

$i_1 R_1=i_2 R_2$(same potential difference)$V=I_1 {\text{R}}_1=I_1 \times \;\frac{\rho l_1}{\pi r_1^2}$Also $V=I_2 R_2=I_2 \times \;\frac{\rho l_2}{\pi r_2^2}$ $\; \therefore l_1 \times \;\frac{\rho l_1}{\pi r_1^2}=I_2 \times \;\frac{\rho l_2}{\pi r_2^2}$ $\;\Rightarrow \;\frac{I_1}{I_2}=\;\frac{ℓ_1}{ℓ_2} \times \;\frac{r_1^2}{r_2^2}$ $\;=\;\frac{3}{4} \times \;\frac{4}{9}=\;\frac{1}{3}$

NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoffs second law is based on conservation of energy.

$S=R_1+R_2$ and $P=\;\frac{R_1 R_2}{R_1+R_2}$ $S=n P \Rightarrow R_1+R_2=\;\frac{n (R_1 R_2) }{ (R_1+R_2) }$ $\Rightarrow (R_1+R_2) ^2=n R_1 R_2$ $\Rightarrow n=\;\frac{R_1^{1+}R_2^{2+}R_1 R_2}{R_1 R_2}$ $n=\;\frac{R_1}{R_2}+\;\frac{R_2}{R_1}+2$Arithmetic mean $>$ Geometric meanMinimum value of $n$ is 4

The mass liberated $m$, electrochemical equivalent of a metal $Z$, are related as $m=Z i t$ $\;\Rightarrow m=3.3 \times 10^{-7} \times 3 \times 2$ $\;=19.8 \times 10^{-7} {\text{kg}}$

Potential gradient along wire, ${\text{K}}=\;\frac{E}{100}$ volt/cmFor battery ${\text{V}}= {\text{E}}^{'}$ - ir, where ${\text{E}}^{'}$ is emf of battery.or ${\text{K}} \times 30= {\text{E}}^{'}- {\text{ir}}$, where current ${\text{i}}$ is drawn from batteryor $\;\frac{E \times 30}{100}= {\text{E}}^{'}+0.5 {\text{i}}$or ${\text{E}}^{'}=\;\frac{30 E}{100}-0.5 i$

$i_g \times G= (i-i_g) S$ $\therefore S=\;\frac{i_g \times G}{i-i_g}=\;\frac{1 \times 0.81}{10-1}=0.09 Ω$

$R_f=n^2 R_1$Here $n=2$ (length becomes twice)$\therefore R_f=4 R_i$New resistance $=400$ of $R_i$ $\therefore \;\text{ Increase }\;=300 \%$

Let $\theta$ be the smallest temperature difference that can be detected by the thermocouple, then$I \times R= (25 \times 10^{-6}) \theta$where ${\text{I}}$ is the smallest current which can be detected by the galvanometer of resistance $R$.$\therefore 10^{-5} \times 40=25 \times 10^{-6} \times \theta$ $\therefore \theta =16^{\circ} C \;\text{. }\;$

$R_p=\;\frac{3 \times 6}{3+6}=\;\frac{18}{9}=2 Ω$ $\therefore V=I R$ $\Rightarrow I=\;\frac{V}{R}=\;\frac{3}{2}=1.5 {\text{A}}$

According to Faraday's first law of electrolysis $m=z \times q$For same $q, \;\; m \propto Z$ $\; \therefore \;\frac{m_{C n}}{m_{Z_n}}=\;\frac{Z_{C u}}{Z_{Z_n}}$ $\;\Rightarrow m_{C u}=\;\frac{Z_{C u}}{Z_{Z_n}} \times m_{Z n}$ $\;=\;\frac{31.5}{32.5} \times 0.13=0.126 {\text{g}}$

We know that $R=\;{V_{\;\text{ruted }\;}^2}/{P_{\;\text{rated }\;}}=\;\frac{(220)^2}{1000}$When this bulb is connected to 110 volt mains supply we get$P=\;\frac{V^2}{R}=\;\frac{(110)^2 \times 1000}{(220)^2}=\;\frac{1000}{4}=250 W$

According to Faraday's first law of electrolysis $m=Z I t \Rightarrow m \propto I t$

The equivalent resistance is ${\text{R}}_{e q}=\;\frac{2 \times R}{2+R}$ $\therefore$ Powder dissipation $P=\;{V^2}/{ {\text{R}}_{e q}}$ $\therefore 150=\;\frac{15 \times 15}{R_{e q}}$ $\therefore {\text{R}}_{e q}=\;\frac{15}{10}=\;\frac{3}{2}$ $\Rightarrow \;\frac{2 R}{2+R}=\;\frac{3}{2}$ $\Rightarrow 4 R=6+3 R$ $\Rightarrow R=6 Ω$

$\theta _n=\;\frac{\theta _i+\theta _c}{2}$

Case $1: P_1=\;\frac{V^2}{R}$ Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $\frac{R}{2}$. These are connected in parallel$\therefore {\text{R}}_{e q}=\;\frac{R {/} 2}{2}=\;\frac{R}{4}$ $\therefore P_2=\;\frac{V^2}{R {/} 4}=4(\;\frac{V^2}{R})=4 P_1$

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer. The same procedure needs to be done if ammeter is to be used as a voltmeter.