Current Electricity Practice Questions

Potential different across all resistor is same So they are in parallel$\;\;\frac{1}{R}=\;\frac{1}{20}++\;\frac{1}{20}+\;\frac{1}{10}$ $\;R_{ {\text{eq}}}=5 {\text{k}} Ω$

$\;\;\text{ Req. }\;=0.5+1+4.5+ (\;\frac{3.6}{9})$ $\;\;\text{ Req. }\;=6+2=8 Ω$ $\; {\text{I}}=\;\frac{8-4}{8}=\;\frac{1}{2} {\text{A}}=0.5 {\text{A}}$ $\; {\text{I}}_1: {\text{I}}_2=\;\frac{1}{3}: \;\frac{1}{6}$ $\; {\text{I}}_1: {\text{I}}_2=2: 1$ $\;\;\text{ and }\; {\text{I}}_1+ {\text{I}}_2=0.5 {\text{A}}$ $\; {\text{I}}_1=\;\frac{2}{3} \times 0.5=\;\frac{1}{3} {\text{A}}$ $\;\;\text{ So }\; \;\frac{1}{3}=\;{ {\text{x}}}/{3} \Rightarrow {\text{x}}=1$

$\;\;\frac{\epsilon }{r+5} \times 5=200 x$ . . . (1)$\;\;\frac{\epsilon \times 15}{r+15}=300 x$ . . . (2)$\;\Rightarrow r=5 Ω$Ans. 5

${\text{R}}_{\;\text{initial }\;}=\;{\rho ℓ}/{ {\text{A}}}=5 Ω$ $\because$ Volume of wire is constant in stretching$\; {\text{V}}_{ {\text{i}}}= {\text{V}}_{ {\text{f}}}$ $\; {\text{A}}_{ {\text{i}}} ℓ_{ {\text{i}}}= {\text{A}}_{ {\text{f}}} ℓ_{ {\text{f}}}$ $\; {\text{A}} ℓ= {\text{A}}^{'}(5 ℓ)$ $\; {\text{A}}^{'}=\;{ {\text{A}}}/{5}$ $\; {\text{R}}_{ {\text{f}}}=\;{\rho ℓ_{ {\text{f}}}}/{ {\text{A}}_{ {\text{f}}}}=\;{\rho (5 ℓ)}/{ (\;{ {\text{A}}}/{5}) }$ $\;=25 (\;{\rho ℓ}/{ {\text{A}}})$ $\;=25 \times 5=125 Ω$

All $9 Ω$ resistances are in parallel$R_{e q}=3 Ω$ $I=\;\frac{6}{3} A=2 A$

$R_{n e t}=\;\frac{m a}{3}+\;\frac{a}{2 m}$ $=a [\;\frac{m}{3}+\;\frac{1}{2 m}-\;{2}/{{\sqrt{6}}}+\;{2}/{{\sqrt{6}}}]$ $=a [ (\sqrt{\;\frac{m}{3}}-\;{1}/{{\sqrt{2 m}}}) ^{2+}\sqrt{\;\frac{2}{3}}]$This will be minimum when$\sqrt{\;\frac{m}{3}}=\;{1}/{{\sqrt{2 m}}}$ $\;\text{ or }\; m=\sqrt{\;\frac{3}{2}}$ $\;\text{ so }\; x=3$

Balanced wheat stone bridge in circuit so there is no current in $5 Ω$ resistor so it can be removed from the circuit. ${\text{R}}_{ {\text{eq}}}=\;\frac{6 \times 12}{6+12}+2$ $=\;\frac{6 \times 12}{18}+2$ ${\text{R}}_{ {\text{eq}}}=6 Ω$ ${\text{I}}=\;{ {\text{V}}}/{ {\text{R}}_{ {\text{eq}}}}=\;\frac{6}{6}=1 {\text{Amp}}$.

$\Rightarrow \;\frac{S}{S+G} i=n K$ $\Rightarrow i=\;\frac{n K(S+G)}{S}$

$\;\frac{1.5 \times 10}{10+\;\frac{r}{2}}=1.2$ $\Rightarrow r=5 Ω$

$V_B-V_A=i \times 2$ $=\;\frac{15}{1+2} \times 2$ $\Rightarrow V_B-V_A=10 \;\text{ volts }\;$

$E \propto I$ $\;\frac{1.2}{1.8}=\;\frac{36}{I^{'}}$ $I^{'}=\;\frac{3}{2} \times 36=54 {cm}$ $\Delta I=I^{'}-I=54-36=18 {cm}$

Resistance $=\tan 45^{\circ}=1 Ω$ $\Rightarrow 1=\;\frac{p I}{A}$ $\Rightarrow p=\;{\pi (1.2 {\text{cm}})^2}/{31.4 {\text{cm}}}=1.44 \times 10^{-1} Ω {\text{cm}}$ $\Rightarrow x=144$

${\text{R}}_{10}=2= {\text{R}}_0(1+\alpha \times 10)$ ${\text{R}}_{30}=3= {\text{R}}_0(1+\alpha \times 30)$On solving$\alpha =0.033 /{ }^{\circ} {\text{C}}$

Let resistance of potentiometers wire is ${\text{R}}$ $i=\;\frac{4}{R+780}$Potential difference across ${\text{AB}}$ $=\;\frac{4 R}{R+780}$Potential difference across AC$=\;\frac{4 R \times 60}{(R+780) \times 300}=\;\frac{4 R}{5(R+780)}$This should be equal to $20 {\text{mV}}$ $\;{4 {\text{R}}}/{5( {\text{R}}+780)}=20 \times 10^{-3}=2 \times 10^{-2}$ $4 {\text{R}}=10^{-1}( {\text{R}}+780)$ $4 {\text{R}}=\;{ {\text{R}}}/{10}+78$ $4 {\text{R}}-\;{ {\text{R}}}/{10}=78$ $\;{39 {\text{R}}}/{10}=78$ ${\text{R}}=20 Ω$

${\text{R}}_1=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{220^2}{100}=484$ ${\text{R}}_2=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{220^2}{60}=484(\;\frac{10}{6})$ ${\text{I}}=\;\frac{220}{484+484 \times \;\frac{10}{6}}$ ${\text{P}}_1= {\text{I}}^2 {\text{R}}_1=14.06 {\text{W}}$

$\;\frac{\epsilon _1 8}{r_1+8}=3 c$ $\;\frac{\epsilon _1 4}{r_1+4}=2 c$ $\Rightarrow \;\frac{2 (r_1+4) }{r_1+8}=\;\frac{3}{2}$ $\Rightarrow r_1=8 Ω$

${\text{R}}_1=\rho \;{ {\text{L}}_1}/{ {\text{A}}_1}$ ${\text{R}}_2=\rho (\;{3 {\text{L}}_1}/{ {\text{A}}_1 / 3})=9 \rho \;{ {\text{L}}_1}/{ {\text{A}}_1}$ $\therefore \;{ {\text{R}}_2}/{ {\text{R}}_1}=9$

Using the conditions of a balanced wheat stone bridge and adding the end correction.$\;\frac{15}{(43+2)}=\;\frac{R}{(102-45)} \Rightarrow R=\;\frac{57}{45} \times 15$ $R=19 Ω$