Current Electricity Practice Questions

$R A B=R$ $R=\;\frac{1}{8}$ (Resistance of one wire) $=\;\frac{1}{8} \rho \;{l}/{\pi \;\frac{d^{2}}{4}}=\;\frac{\rho l}{2 \pi d^{2}}$Resistance of copper wire of length $2 l$ and diameter $x=R$ $\rho \;{2 l}/{\pi \;\frac{x^{2}}{4}}=R$ $\;\frac{8 \rho l}{\pi x^{2}}=\;\frac{\rho l}{2 \pi d^{2}}$ $16 d^{2}=x^{2}$ $x=4 d$

Let the current is ${\text{i}}$Using Kirchhoff's law$i R+\;\frac{i}{2} R+\;\frac{i}{4} R+\;\frac{i}{8} R=3$ $i=\;\frac{3 \times 8}{15}=\;\frac{8}{5} A$So $a=8$

$i=A \times j$ $=\pi (R^{2-}\;\frac{R^2}{4}) j$ $=\;\frac{3 \pi R^2}{4} \times j$ $=\;\frac{3 \pi \times (4 \times 10^{-3}) ^2}{4} \times 4 \times 10^6$ $=48 \pi$

$\Rightarrow I=\;\frac{5}{2.5}=2 A$

$\;{ {\text{R}}_{ {\text{X}}}}/{ {\text{R}}_{ {\text{Y}}}}=\;{ℓ_{ {\text{X}}}}/{ℓ_{ {\text{Y}}}}$When wire is stretched to double of its length, then resistance becomes 4 times${\text{R}}_{ {\text{W}}}=4 {\text{R}}_{ {\text{X}}}=2 {\text{R}}_{ {\text{Y}}}$ $\;{ {\text{R}}_{ {\text{X}}}}/{ {\text{R}}_{ {\text{Y}}}}=\;\frac{1}{2}$So. $\;\frac{ℓ_x}{ℓ_y}=\;\frac{1}{2}$

$i=A \times j$ $=\pi (R^{2-}\;\frac{R^2}{4}) j$ $=\;\frac{3 \pi R^2}{4} \times j$ $=\;\frac{3 \pi \times (4 \times 10^{-3}) ^2}{4} \times 1.0 \times 10^6$ $=12 \pi$

$\therefore E=I \times (\;\frac{20}{4})=\;\frac{25}{(30+20)} \times (\;\frac{20}{4})$ $=\;\frac{1}{2} \times 5=2.5$ volts $=\;\frac{25}{10}$ volts

Even if the radius of wire is doubled, the balancing point would not change as $\;\frac{x}{l-x}=\;\frac{R_1}{R_2}$, which is not including a term of area.

Initially $5 Ω$ and $5 Ω$ are in series and then in parallel with $10 Ω$ this pattern continues thus ${\text{R}}_{\;\text{net }\;}=5 Ω$

At balancing point, we know that emf is proportional to the balancing length. i.e.,emf $\propto$ balancing lengthNow, let the emf's be $3 \epsilon$ and $2 \epsilon$.$\Rightarrow 3 \epsilon = {\text{k}}(75) . . . . .(1)$and $2 \epsilon = {\text{k}}( {\text{I}}) . . . . . .(2)$ $\Rightarrow {\text{I}}=50 {\text{cm}}$ $\Rightarrow$ Difference is $(75-50) {\text{cm}}=25 {\text{cm}}$.

${\text{I}}=\;{2 {\text{E}}}/{ {\text{r}}_1+ {\text{r}}_2+ {\text{R}}}$ ${\text{IR}}= {\text{E}}- {\text{Ir}}_2$ ${\text{I}}( {\text{R}}+ {\text{r}}_2)= {\text{E}}$ ${\text{I}}=\;{ {\text{E}}}/{ {\text{R}}+ {\text{r}}_2}$ $\;{2 {\text{E}}}/{ {\text{r}}_1+ {\text{r}}_2+ {\text{R}}}=\;{ {\text{E}}}/{ {\text{R}}+ {\text{r}}_2}$ $2 {\text{R}}+2 {\text{r}}_2= {\text{r}}_1+ {\text{r}}_2+ {\text{R}}$ ${\text{R}}= {\text{r}}_1- {\text{r}}_2$

$300=I^2 R \times 15$ $\Rightarrow R=5 Ω$Now $I_2^2 R t_2$ $=9 \times 5 \times 10$ $=450 {\text{J}}$

According to given circuit diagram When switch is open, then combination $R_{1}$ and $R_{2}$ will be in series and also combination $R_{3}$ and $R_{4}$ will be in series and these branches will be in parallel. $\therefore$ Equivalent resistance $(R_{\;\text{eq }\;})=\;\frac{3 R \cdot 3 R}{3 R+3 R}=\;\frac{9 R^{2}}{6 R}=\;\frac{3}{2} R$When switch is closed, then combination $R_{1}$ and $R_{3}$ will be in parallel and also, combination $R_{2}$ and $R_{4}$ will be in parallel. After that, both will be in series.$\therefore$ Equivalent resistance, $({R^{'}}_{\text{eq}}) \;\; =\;\frac{R \cdot 2 R}{R+2 R}+\;\frac{R \cdot 2 R}{R+2 R}=\;\frac{2 R^{2}}{3 R}+\;\frac{2 R^{2}}{3 R}$ $\;\; =\;\frac{2 R}{3}+\;\frac{2 R}{3}=\;\frac{4 R}{3}$Now, dividing Eq. (i) by Eq. (ii), we get ${c} \;{R_{\;\text{eq }\;}}/{R_{\;\text{eq }\;}{ }^{'}}=\;\frac{3 \frac{R}{2}}{4 \;\frac{R}{3}}=\;\frac{9}{8}$ $x=9$

Given, $\;\; i=\alpha _{0} t+\beta t^{2}$ where, $\alpha _{0}=20 {A} / {s}, \beta =8 {As}^{2}$ We know that, $\;\; i=\;\frac{d q}{d t}$ $\Rightarrow \;\;\frac{d q}{d t}=i=\alpha _{0} t+\beta t^{2}=20 t+8 t^{2}$ $\Rightarrow \;d q=(20 t+8 t^{2}) d t$ On integrating both sides, we get $∫_{0}^{q} d q=∫_{0}^{15}(20 t+8 t^{2}) d t$ q=\[;\frac{20 t^{2}}{2}+;\frac{8 t^{3}}{3}]_{0}^{15}=10 ×(15)^{2}+;\frac{8}{3} ×(15)^{3}$$q=11250 {C}$

The given value of resistances are $2 \Omega, 4 \Omega, 6 \Omega$ and $8 \Omega$. The required value of combination is $46 / 3 \Omega$. In order to achieve the above mentioned values of resistance from given resistances, we will connect 2Ω and 4Ω resistance in parallel, then join 6 Ω and 8 Ω resistance in series with the combination. The circuit diagram for connection is shown below. $\therefore \text{R}_{{\text{eq}}}=(2 \| 4)+6+8=\frac{2 \times 4}{2+4}+14=\frac{46}{3} \Omega$ Thus, resistance of 2Ω and 4Ω are in parallel with 6Ω and 8Ω in series combination.

Applying KCL at point P, $\frac{V-0}{6}+\frac{V-90}{5}+\frac{V-140}{20}=0$ $\Rightarrow10 {V}+12 {V}-1080+3 {V}-420=0$ $\Rightarrow{V}=60$ $\therefore$ current in $6 \Omega={{V}-0}{6}=10 {A}$ Hence option 3.

At the balanced condition of the Wheatstone bridge, $\;\frac{R}{S}=\;\frac{L_{1}}{L-L_{1}}$ $\Rightarrow \;\; \;\frac{12}{6}=\;\frac{x}{72-x}$ $\Rightarrow \;\; x=48 {cm}$ $\therefore$ The galvanometer jockey is placed at $P$ on $A B$ at a distance of 48 ${cm}$ from the $A$. So, the value of the $x$ to the nearest integer is 48 .

$16={R}_{0}[1+\alpha (15-{T}_{{o}})]$ $20={R}_{0}[1+\alpha (100-{T}_{{o}})]$ Assuming ${T}_{{o}}=0°{C}$, as a general convention. $\Rightarrow\frac{16}{20}=\frac{1+\alpha \\times 15}{1+\alpha \\times 100}$ $\Rightarrow\alpha =0.003{ }°{C}^{-1}$