Current Electricity Practice Questions

${\text{H}}_1=\;{ {\text{V}}^2}/{( {\text{R}} / 2)} {\text{t}}=\;{2 {\text{V}}^2}/{ {\text{R}}}$ . . . (1)${\text{H}}_2=\;{ {\text{V}}^2}/{2 {\text{R}}} {\text{t}}$ $\;{ {\text{H}}_1}/{ {\text{H}}_2}= (\;{2 {\text{V}}^2 {\text{t}}}/{ {\text{R}}}) \times \;{2 {\text{R}}}/{ {\text{V}}^2 {\text{t}}}=\;\frac{4}{1}$

${\text{R}}=\rho \;{ℓ}/{ {\text{A}}}$, the cross-sectional area is $\pi ( {\text{b}}^{2-} {\text{a}}^2)$ $\; {\text{R}}=\rho \;{ℓ}/{\pi ( {\text{b}}^{2-} {\text{a}}^2) }=\;\frac{2.4 \times 10^{-8} \times 3.14}{3.14 \times (4^{2-}2^2) \times 10^{-6}}$ $\;=2 \times 10^{-3} Ω$ $\;\to {\text{n}}=2$

$\;\;\frac{2}{3}=\;\frac{\;\frac{x}{x+1}}{x}$ $\;\Rightarrow \;\frac{2}{3}=\;\frac{1}{x+1}$ $\;\Rightarrow \;\; x=0.5=\;\frac{1}{2}$ $\;n=2$

$\; {\text{E}}_{ {\text{eq}}}=\;\frac{\;\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\;\frac{1}{6}}$ $\; {\text{E}}_{ {\text{eq}}}=6 {\text{V}}$ $\; {\text{r}}_{ {\text{eq}}}=2 Ω$ $\; {\text{R}}=4 Ω$So, $i=\;\frac{6}{2+4}=1 {\text{A}}$

Method-I:$\; {\text{R}}_{ {\text{eq}}}=2+\;{5 {\text{R}}}/{5+ {\text{R}}}=\;{10+7 {\text{R}}}/{5+ {\text{R}}}$ $\; {\text{i}}=\;{3}/{ {\text{R}}_{ {\text{eq}}}}=\;{3(5+ {\text{R}})}/{10+7 {\text{R}}}$ $\; {\text{i}}_1=\;\frac{2}{5}, {\text{i}}_2=\;{2}/{ {\text{R}}}$ $\; {\text{i}}= {\text{i}}_1+ {\text{i}}^2$ $\;\;{3(5+ {\text{R}})}/{10+7 {\text{R}}}=\;\frac{2}{5}+\;{2}/{ {\text{R}}}=\;{2(5+ {\text{R}})}/{5 R}$ $\;15 {\text{R}}(5+ {\text{R}})=2(5+ {\text{R}})(10+7 {\text{R}})$ $\;75 {\text{R}}+15 {\text{R}}^2=2 (50+35 {\text{R}}+10 {\text{R}}+2 {\text{R}}^2)$ $\;15 {\text{R}}^{2+}75 {\text{R}}=14 {\text{R}}^{2+}90 {\text{R}}+100$ $\; {\text{R}}^{2-}15 {\text{R}}-100=0$ $\; {\text{R}}=\;{15 {\sqrt{225 \times 1 \times 100}}}/{2}$ $\;=\;{15 ± {\sqrt{625}}}/{2}=\;\frac{15 ± 25}{2}$ $\;R=20 Ω$Method-II: Given potential across $5 Ω$ and voltmeter is $2 {\text{V}}$. To find resistance ${\text{R}}$ of voltmeter.Let current in $5 Ω$ be ${\text{i}}_1$, and in ${\text{R}} {\text{i}}_2$.${\text{i}}_1=\;\frac{2}{5} \;\text{ and }\; {\text{i}}_2=\;{2}/{ {\text{R}}}$ ${\text{V}}$ across $2 Ω$ will be 1 volt and ${\text{i}}=\;\frac{1}{2} {\text{A}}$.Using junction law: ${\text{i}}= {\text{i}}_1+ {\text{i}}_2$ $\;\;\frac{1}{2}=\;\frac{2}{5}+\;{2}/{ {\text{R}}}$ $\;\;{2}/{ {\text{R}}}=\;\frac{1}{2}-\;\frac{2}{5}=\;\frac{1}{10}$ $\; {\text{R}}=20 Ω$

$\; {\text{R}}_{ {\text{eq}}}=4 Ω$ $\; {\text{i}}=\;\frac{8}{4}=2 {\text{A}}$ $\; {\text{i}}_1=\;\frac{2 \times 3}{3+6}=\;\frac{2}{3} {\text{A}}$ $\; {\text{i}}_2=\;\frac{2 / 3}{2}=\;\frac{1}{3} {\text{A}}$

Wheat stone bridge is in balanced condition.$\;\;\frac{1}{R_{e q}}=\;\frac{1}{4 R}+\;\frac{1}{8 R}$ $\;R_{e q}=\;\frac{8 R}{3}$

This question is based on the conceptual clarity that we should connect ammeter in series and voltmeter in parallel to measure current and potential difference, respectivelyAlso, when we use a galvanometer to create an ammeter, shunt resistance should be very small and should be in parallel.When we create a voltemeter shunt should be large and in series with galvanometer.All these criteria are satisfied in option (2)

$i=\;\frac{2 \epsilon }{5+2 r}$(1) $\;\;$ i $=\;{\epsilon }/{{ {\text{r}}}/{2}+5}$ . . . (2)Equating (1) and (2)$\;\;{2 \epsilon }/{5+2 {\text{r}}}=\;{\epsilon }/{{ {\text{r}}}/{2}+5} \Rightarrow {\text{r}}+10=5+2 {\text{r}}$ $\; {\text{r}}=5 \;\text{ Ans. }\; 5$

$\; {\text{H}}=\;{ {\text{V}}^2}/{ {\text{R}}} \times {\text{t}}$ $\;\;{ {\text{H}}_1}/{ {\text{H}}_2}=\;{\;{ {\text{V}}^2 {\text{t}}}/{ {\text{R}}}}/{{ {\text{V}}^2 {\text{t}}}/{3 {\text{R}}}}=3: 1$

$\;R_{\max } \Rightarrow \;\text{ \in series }\; \Rightarrow 10 R=10 \times 10=100 Ω$ $\;R_{\max } \Rightarrow \;\text{ \in parallel }\;=\;\frac{R}{10}=\;\frac{10}{10}=1 Ω$ $\;\;\frac{R_{\max }}{R_{\min }}=\;\frac{100}{1}=100 \;\text{ Ans. }\;$ $\;R_{\min } \Rightarrow \;\frac{100}{1}=100 \;\text{ Ans. }\;$

$\;\rho =3 \times 10^{-7} Ω- {\text{cm}}$ $\; {\text{R}}=\rho \cdot \;{1}/{ {\text{A}}}$ $\;=\;{3 \times 10^{-7} \times (10^{-2} {\text{m}}) }/{ (100 \times 10^{-4} {\text{m}}^2) }=3 \times 10^{-7}$

$\;R=\;\frac{\rho l}{A}$ $\;R^{'}=\;\frac{\rho \times 1.2 l}{0.96 A}=\;\frac{10}{8} \times R=1.25 R$It means $25 \%$ increase in Resistance.

$\; {\text{R}}_{ {\text{eq}}}=\;\frac{6}{2}=3 Ω$ $\; {\text{I}}=\;{ {\text{V}}}/{ {\text{R}}}=\;\frac{9}{3}=3 {\text{A}}$ $\; {\text{V}}_{ {\text{A}}}+(1.5)(2)-(1.5) 4= {\text{V}}_{ {\text{B}}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{B}}}=3 {\text{V}}$

$\tau = {\text{mB}}\; {\text{A}}=\;\text{ area of coil }\;$ ${\text{K}} \theta = {\text{IANB}}\; {\text{B}}=\;\text{ magnetic field }\;$ $\;{\theta }/{ {\text{I}}}=\;{ {\text{ANB}}}/{ {\text{K}}}\;\;\text{ Currect senstivity }\;$ $1.25 (\;{\theta }/{ {\text{I}}}) _2= (\;{\theta }/{ {\text{I}}}) _1\;. . . . . . . . . .(1)$ $1.25 [\;{ {\text{AN}}_2 {\text{B}}}/{ {\text{K}}}] = [\;{ {\text{AN}}_1 {\text{B}}}/{ {\text{K}}}]$ $1.25=\;{ {\text{N}}_1}/{ {\text{N}}_2}=\;\frac{5}{4}\;. . . . . .(2)$ $\Rightarrow {\text{R}}=\;{\delta ℓ}/{ {\text{a}}}=\;\text{ const. }\;\;$ $\Rightarrow ℓ= {\text{a}}$Voltage sensitivity $=\;\frac{\theta }{V}=\;\frac{\theta }{ℝ}=\;{\;\text{ Current sensitivity }\;}/{ {\text{R}}}$ ${\text{R}}=$ constantVoltage sensitivity $\propto$ current sensitivity

Voulme remain same$\; {\text{A}} ℓ= {\text{A}}^{'} ℓ^{'}$ $\; {\text{A}}^{'}=\;{ {\text{A}} ℓ}/{ℓ^{'}}=\;{ {\text{A}} ℓ}/{ℓ / 4}$ $\; {\text{A}}^{'}=4 {\text{A}}$ $\; {\text{L}}^{'}=\;{ {\text{L}}}/{4}$ $\;\;{ {\text{R}}^{'}}/{ {\text{R}}^{'}}=\;{\rho { {\text{L}}^{'}}/{ {\text{A}}^{'}}}/{{ {\text{L}}}/{ {\text{A}}}}$ $\; {\text{R}}^{'}= {\text{R}} [\;{ {\text{L}}^{'}}/{ {\text{A}}} \times \;{ {\text{A}}}/{ {\text{L}}}]$ $\; {\text{R}}^{'}=160 [\;{ {\text{L}}}/{4 {\text{L}}} \times \;{ {\text{A}}}/{4 {\text{A}}}]$ $\; {\text{R}}^{'}=10 Ω$

Both $4 Ω$ resistance gets short.Remove the resistors that have no current.$\; {\text{R}}_{ {\text{eq}}}=3+(2 \| 2)+6$ $\; {\text{R}}_{ {\text{eq}}}=3+1+6$ $\; {\text{R}}_{ {\text{eq}}}=10 Ω$

Power dissipation, ${\text{P}}= {\text{I}}^2 {\text{R}}$(A) ${\text{R}}_{ {\text{eq}}}=\;{ {\text{R}}}/{2}+ {\text{R}}=\;{3 {\text{R}}}/{2}$(B) $R_{\;\text{eq }\;}=\;\frac{(2 R)(R)}{2 R+R}=\;\frac{2 R}{3}$(C) ${\text{R}}_{ {\text{eq}}}=\;{ {\text{R}}}/{3}$(D)$\;R_{\;\text{eq }\;}=3 R$ $\;R_D > R_A > R_B > R_C$Since, $P \propto R_{\;\text{eq }\;}$ ${\text{P}}_{ {\text{D}}} > {\text{P}}_{ {\text{A}}} > {\text{P}}_{ {\text{B}}} > {\text{P}}_{ {\text{C}}}$