Current Electricity Practice Questions

$\; (I-I_g) R=I_g G$ $\;(3-0.02) \times R=0.02 \times G \Rightarrow R=\;\frac{30}{149}$ $\;\Rightarrow 149=\;\text{ Required }\; X$

For the wire bent into an equilateral triangle, each side has a resistance $\;\frac{R}{3}$.$R_{e q}=\;\frac{ (\frac{2 R}{3}) (\frac{R}{3}) }{\frac{2 R}{3}+\;\frac{R}{3}}=\;\frac{2 R}{9}=R_1$ (lets say)For the wire bent into a square, each side has a resistance $\;\frac{R}{4}$.$R_{e q}=\;\frac{ (\frac{3 R}{4}) (\frac{R}{4}) }{\frac{3 R}{4}+\;\frac{R}{4}}=\;\frac{3 R}{16}=R_3 \;\text{ (lets say) }\;$ $\Rightarrow \;\; \;\frac{R_1}{R_3}=\;\frac{\;\frac{2 R}{9}}{\frac{3 R}{16}}=\;\frac{32}{27}$

Clearty, $R_1=\;\frac{R}{6}$So $R_{A B}=R_1|2 R_1\| 2 R_1$ $\;\Rightarrow \;\frac{1}{R_{A B}}=\;\frac{1}{R_1}+\;\frac{1}{2 R_1}+\;\frac{1}{2 R_1}=\;\frac{4}{2 R_1}=\;\frac{2}{R_1}$ $\;\Rightarrow R_{A B}=\;\frac{R_1}{2}=\;\frac{R}{12}$

Equivalent resistance across $A C$ is$\;R_1=2 R \| R=\;\frac{2 R \times R}{3 R}=\;\frac{2 R}{3}$ $\;\Rightarrow R_1=2 Ω$

Let $R$ be total resistance across ends of wire, then$R_{ {\text{eq}}}\;=\;\frac{R}{4}=\;\frac{\rho ℓ}{4 A}=\;\frac{2 \times 10^{-6} \times 25}{4 \times 5 \times 10^{-6}}$ $\;=2.5 Ω$None matches the options.

$I=\;{V}/{{\sqrt{R^{2+}\omega ^2 L^2}}}$ $V_R=\;{R}/{{\sqrt{R^{2+}\omega ^2 L^2}}} V$

$\; {\text{R}}= {\text{R}}_0(1+\alpha \Delta {\text{T}})$ $\;\;{\Delta {\text{R}}}/{ {\text{R}}_0}=\alpha \Delta {\text{T}}$Case-I$\;0{ }^{\circ} {\text{C}} \longrightarrow 100^{\circ} {\text{C}}$ $\;\;\frac{10.2-10}{10}=\alpha (100-0)$ . . . (1)Case-II$\;0{ }^{\circ} {\text{C}} \longrightarrow {\text{t}}^{\circ} {\text{C}}$ $\;\;\frac{10.95-10}{10}=\alpha ( {\text{t}}-0)$ . . . (2)$\;\Rightarrow \;{ {\text{t}}}/{100}=\;\frac{0.95}{0.2}=475^{\circ} {\text{C}}$ $\; {\text{t}}=475+273=748 {\text{K}}$

$\;R_{e q}=4000 Ω$ $\;i=\;\frac{4}{4000}=\;\frac{1}{1000} {\text{A}}$ $\;V_0=i . R=\;\frac{1}{1000} \times 500=0.5 {\text{V}}$

As specific resistance does not depends on dimension of wire so, it will not change.

We know ${\text{R}}=\;{\rho l}/{ {\text{A}}}, {\text{R}} \propto \;{l}/{ {\text{r}}^2}$As we starch the wire, its length will increase but its radius will decrease keeping the volume constant$\; {\text{V}}_{ {\text{i}}}= {\text{V}}_{ {\text{f}}}$ $\;\pi {\text{r}}^2 l=\pi \;{ {\text{r}}^2}/{4} l_{ {\text{f}}}$ $\;l_{ {\text{f}}}=4 l$ $\;\;{ {\text{R}}_{\;\text{new }\;}}/{ {\text{R}}_{\;\text{old }\;}}= (\;{4 l}/{ {\text{r}}^2} \;{ {\text{r}}^2}/{l}=16.$ $\; \;\; {\text{R}}_{\;\text{new }\;}=16 {\text{R}}$ $\; \therefore \;\; {\text{x}}=16$

Let potential gradient be $\lambda$.$\; \therefore {\text{i}} \times 10=\lambda \times 500=\epsilon - {\text{ir}}_{ {\ s}}$ $\;\Rightarrow 500 \lambda =\epsilon -50 \lambda {\text{r}}_{ {\ s}}$Also,$\; {\text{i}}^{'} \times 1=\lambda \times 400=\epsilon - {\text{i}}^{'} {\text{r}}_{ {\ s}}$ $\;\Rightarrow 400 \lambda =\epsilon -400 \lambda {\text{r}}_{ {\ s}}$ $\; \therefore 100 \lambda =350 \lambda {\text{r}}_{ {\ s}} \Rightarrow {\text{r}}_{ {\ s}}=\;\frac{10}{35} \approx 0.3 Ω$Hence option (4)

$\; {\text{i}}= {\text{K}} \theta$ $\;\;{2}/{ {\text{G}}+ {\text{R}}}= {\text{K}} \theta$ $\;\Rightarrow \;\frac{1}{\theta }=\;{( {\text{G}}+ {\text{R}}) {\text{K}}}/{2}= {\text{R}} (\;{ {\text{K}}}/{2}) +\;{ {\text{KG}}}/{2}$ $\;\;\text{ Slope }\;=\;{ {\text{K}}}/{2}=\;\frac{1}{4} \Rightarrow {\text{K}}=0.5=5 \times 10^{-1} {\text{A}}$

$R\;=\;\frac{V}{I_g}-R_g=\;\frac{100}{5 \times 10^{-3}}-50$ $\;=20000-50$ $\;=19950 Ω$

${\text{i}}=\;\;{ {\text{E}}_{\;\text{eq }\;}}/{ {\text{r}}_{\;\text{eq }\;}}=\;\frac{8 \times 5}{8 \times 0.2}$ $\;I=25 A$ $\;V=E-i r$ $\;=5-0.2 \times 25$ $\;=0$

Given ${\text{G}}=10 Ω$ $\; {\text{I}}_{ {\text{g}}}=3 {\text{mA}}$ $\; {\text{I}}=8 {\text{A}}$In case of conversion of galvanometer into ammeter. We have ${\text{Ig}}_{ {\text{g}}} {\text{G}}= ( {\text{I}}- {\text{I}}_{ {\text{g}}}) {\text{S}}$ $\;S=\;\frac{I_g G}{I-I_g}$ $\;S=\;\frac{ (3 \times 10^{-3}) 10}{8-0.003}$ $=3.75 \times 10^{-3} Ω$

Series : $\; {\text{R}}_{ {\text{eq}}}= {\text{R}}_1+ {\text{R}}_2$ $\;2 {\text{R}} (1+\alpha _{ {\text{eq}}} \Delta \theta ) = {\text{R}} (1+\alpha _1 \Delta \theta )$ $+ {\text{R}} (1+\alpha _2 \Delta \theta )$ $\;2 {\text{R}} (1+\alpha _{ {\text{eq}}} \Delta \theta ) =2 {\text{R}}$ $+ (\alpha _1+\alpha _2) {\text{R}} \Delta \theta$ $\;\alpha _{ {\text{eq}}}=\;\frac{\alpha _1+\alpha _2}{2}$Parallel : $\;\;{1}/{ {\text{R}}_{ {\text{eq}}}}=\;{1}/{ {\text{R}}_1}+\;{1}/{ {\text{R}}_2}$ $\;\;{1}/{{ {\text{R}}}/{2} (1+\alpha _{ {\text{eq}}} \Delta \theta ) }=\;{1}/{ {\text{R}} (1+\alpha _1 \Delta \theta ) }$ $+\;{1}/{ {\text{R}} (1+\alpha _2 \Delta \theta ) }$ $\;\;{2}/{1+\alpha _{ {\text{eq}}} \Delta \theta }=\;\frac{1}{1+\alpha _1 \Delta \theta }$ $+\;\frac{1}{1+\alpha _2 \Delta \theta }$ $\;\;{2}/{1+\alpha _{ {\text{eq}}} \Delta \theta }=\;\frac{1+\alpha _2 \Delta \theta +1+\alpha _1 \Delta \theta }{ (1+\alpha _1 \Delta \theta ) (1+\alpha _2 \Delta \theta ) }$ $\;2 [ (1+\alpha _1 \Delta \theta ) (1+\alpha _2 \Delta \theta ) ]$ $\;= [2+ (\alpha _1+\alpha _2) \Delta \theta ] [1+\alpha _{ {\text{eq}}} \Delta \theta ]$ $\;2 [1+\alpha _1 \Delta \theta +\alpha _2 \Delta \theta +\alpha _1 \alpha _2 \Delta \theta ]$ $= \;2+2 \alpha _{ {\text{eq}}} \Delta \theta + (\alpha _1+\alpha _2) \Delta \theta$ $+\alpha _{ {\text{eq}}} (\alpha _1+\alpha _2) \Delta \theta ^2$ Neglecting small terms $\;2+2 (\alpha _1+\alpha _2) \Delta \theta =2+2 \alpha _{ {\text{eq}}} \Delta \theta$ $+ (\alpha _1+\alpha _2) \Delta \theta$ $\; (\alpha _1+\alpha _2) \Delta \theta =2 \alpha _{ {\text{eq}}} \Delta \theta$ $\alpha _{ {\text{eq}}}=\;\frac{\alpha _1+\alpha _2}{2}$

${\text{i}}_{ {\text{g}}}=\;\frac{10}{400+100}=20 \times 10^{-3} {\text{A}}$For ammeterLet shunt resistance $= {\text{S}}$ $\;i_g R= (i-i_g) S$ $\;20 \times 10^{-3} \times 100=10 S$ $\;S=20 \times 10^{-2} Ω$