Current Electricity Practice Questions

$\;I_S S=I_g G$ $\;\;\frac{95}{100} I S=\;\frac{5 I}{100} G$ ${\text{S}}=\;{ {\text{G}}}/{19}$ ${\text{R}}_{ {\text{A}}}=\;{ {\text{SG}}}/{ {\text{S}}+ {\text{G}}}=\;{\;{ {\text{G}}^2}/{19}}/{{20 {\text{G}}}/{19}}$ ${\text{R}}_{ {\text{A}}}=\;{ {\text{G}}}/{20}$

Rated power & voltage gives resistance$\; {\text{R}}=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{(200)^2}{50}=\;\frac{40000}{50}$ $\; {\text{R}}=800$ $\; {\text{P}}=\;{ ( {\text{V}}_{\;\text{applied }\;}) ^2}/{ {\text{R}}}=\;\frac{(100)^2}{800}$ $\; {\text{P}}=12.5 \;\text{ watt }\;$Hence option 1 is correct.

${\text{i}}\;=\;\frac{3}{1+2}=1 {\text{A}}$ ${\text{v}}\;= {\text{E}}- {\text{ir}}$ $\;=3-1 \times 1=2 {\text{V}}$

${\text{R}}_{ {\text{eq}}}=2 Ω+2 Ω+1 Ω=5 Ω$ ${\text{i}}=\;{ {\text{V}}}/{ {\text{R}}_{ {\text{eq}}}}=\;\frac{10}{5}=2 {\text{A}}$ $\;\text{ Current \in resistance }\; {\text{R}}_3\;=2 \times (\;\frac{4}{4+4})$ $\;=2 \times \;\frac{4}{8}$ $\;=1 {\text{A}}$

$\;\;\frac{v^2}{R}=W$ ........(i) $\;\;\frac{v^2}{\frac{1}{2} (\;\frac{R}{2}) }=W^{'}$ ............(ii) From (i) & (ii), we get $W^{'}=4 W$

$\;\;\text{ Power }\;(P)=V . I$ $\;\Rightarrow 110=(220)(I)$ $\;\Rightarrow I=0.5 {\text{A}}$ $\;\;\text{ Now, }\; I=\;{ {\text{n}} \cdot {\text{e}}}/{ {\text{t}}}$ $\;\Rightarrow 0.5= (\;{ {\text{n}}}/{ {\text{t}}}) (1.6 \times 10^{-19})$ $\;\Rightarrow \;{ {\text{n}}}/{ {\text{t}}}=\;\frac{0.5}{1.6 \times 10^{-19}}$ $\;\Rightarrow \;{ {\text{n}}}/{ {\text{t}}}=31.25 \times 10^{17}$

$\;\;{ {\text{i}}_1}/{ {\text{i}}_2}=\;\frac{10}{5}=\;\frac{2}{1}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{ {\text{i}}_1^2 {\text{R}}_1}/{ {\text{i}}_2^2 {\text{R}}_2}= (\;\frac{2}{1}) ^2 \times \;\frac{5}{10}=\;\frac{2}{1}$

In case of balanced Wheatstone Bridge$\;\;{ {\text{V}}_{ {\text{AB}}}}/{ {\text{V}}_{ {\text{AD}}}}=\;{ {\text{V}}_{ {\text{BC}}}}/{ {\text{V}}_{ {\text{CD}}}} \Rightarrow \;{12}/{6+ {\text{x}}}=\;\frac{0.5}{0.5}$ $\; {\text{x}}=6 Ω$

${\text{i}} \propto \theta$ (angle of deflection) $\; \therefore \;\frac{i_2}{i_1}=\;\frac{\theta _2}{\theta _1} \Rightarrow \;{i_2}/{200 \mu {\text{A}}}$ $=\;\frac{\pi / 10}{\pi / 3}=\;\frac{3}{10}$ $\;\Rightarrow i_2=60 \mu {\text{A}}$

$\; {\text{G}}=200 Ω$ $\; {\text{i}}_{ {\text{g}}}=20 \mu {\text{A}}$ $\; {\text{i}}= {\text{i}}_{ {\text{g}}} (\;{ {\text{G}}}/{ {\text{S}}}+1)$ $\;\Rightarrow 20 \times 10^{-3}=20 \times 10^{-6} (\;{200}/{ {\text{S}}}+1)$ $\;\Rightarrow \;{200}/{ {\text{S}}}=999$ $\;\Rightarrow {\text{S}} \approx 0.2 Ω$

$\Rightarrow {\text{R}}_{ {\text{eq}}}=6 Ω+5 Ω+8 Ω=19 Ω$

Let ${\text{x}}=$ current $/$ division After applying shunt Now $5 {\text{x}} \times {\text{G}}=20 {\text{x}} \times 24$ $\; {\text{G}}=4 \times 24$ $\; {\text{G}}=96 Ω$

$\; {\text{P}}=\;{ {\text{v}}^2}/{ {\text{R}}}, {\text{R}}=\;{\rho ℓ}/{ {\text{A}}}$ $\; {\text{P}} \propto \;\frac{1}{ℓ}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{ {\text{t}}_2}/{ {\text{t}}_1}=\;\frac{15}{20}=\;\frac{ℓ_2}{ℓ_1}$ $\;ℓ_2=\;\frac{3}{4} ℓ_1$

${\text{R}}_{\;\text{heater }\;}=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{(100)^2}{1000}=10 Ω$For heater ${\text{P}}=\;{ {\text{V}}^2}/{R} \Rightarrow {\text{V}}=\sqrt{ {\text{PR}}}$ $\; {\text{V}}={\sqrt{62.5 \times 10}}$ $\; {\text{V}}=25 {\text{v}}$ $\; {\text{i}}_1=\;\frac{75}{10}=7.5 {\text{A}}, {\text{i}}_{ {\text{H}}}=\;\frac{25}{10}=2.5 {\text{A}} .$ $\; {\text{i}}_{ {\text{R}}}= {\text{i}}_1- {\text{i}}_{ {\text{H}}}=5$ $\; {\text{V}}= {\text{IR}}$ $\; {\text{R}}=\;\frac{25}{5}=5 Ω$

Given thatCurrent $I=I_0+\beta t$ $\;I_0=20 {\text{A}}$ $\;\beta =3 {\text{A}} / {\ s}$ $\; {\text{I}}=20+3 {\text{t}}$ $\;\;{ {\text{dq}}}/{ {\text{dt}}}=20+3 {\text{t}}$ $\;∫_{0}^{ {\text{q}}} {\text{dq}}=∫_{0}^{20}(20+3 {\text{t}}) {\text{dt}}$ $\; {\text{q}}=∫_{0}^{20} 20 {\text{dt}}+∫_{0}^{20} 3 {\text{tdt}}$ \; {\text{q}}= \[20 {\text{t}}+;{3 {\text{t}}^2}/{2}]$ _0^{20}=1000 {\text{C}}$

$\;\;\text{ Resistance of each part }\;=\;{ {\text{R}}}/{5}$ $\;\;\text{ Total resistance }\;=\;\frac{1}{5} \times \;{ {\text{R}}}/{5}=\;{ {\text{R}}}/{25}$

$\; {\text{R}}_{\;\text{eq }\;}=\;\frac{4}{3} Ω$ $\; \therefore {\text{P}}=\;{ {\text{V}}^2}/{ {\text{R}}_{\;\text{eq }\;}}=\;\frac{4}{4 / 3}=3 {\text{W}}$

First case $\;\frac{2}{40}=\;\frac{X}{60} \Rightarrow X=3 Ω$In second case ${\text{X}}^{'}=\;\frac{2 \times 3}{2+3}=1.2 Ω$ $\;\;\frac{2}{ℓ}=\;\frac{1.2}{100-ℓ}$ $\;200-2 ℓ=1.2 ℓ$ $\;ℓ=\;\frac{200}{3.2}=62.5 {\text{cm}}$Balance length changes by $22.5 {\text{cm}}$