Current Electricity Practice Questions

According to given circuit diagram, As we know that, parallel equivalent resistance, ${1}/{\text{R}_{{\text{eq}}}}={1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}+...$ and series equivalent resistance, $\text{R}_{{\text{eq}}}=\text{R}_{1}+\text{R}_{2}+\text{R}_{3}+...$ Let the net resistance across a and b be R' $\therefore \frac{1}{\text{R}'}=\frac{1}{2+2+6}+\frac{1}{4+6}$ ${a}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}$ R' = 5 Ω Hence, total resistance, $\text{R}=\text{R}'+1=5+1=6 \Omega$ According to current division rule, current in upper branch, $\text{I}_{1}=\text{I} .\frac{4+6}{(2+2+6)+(4+6)}$ $=1 .\frac{10}{20}=\frac{1}{2}=\frac{1}{2} .\frac{\text{V}}{\text{R}}=\frac{1}{2} \times \frac{12}{6}=1 {\text{A}}$ Again, according to current division rule, current in 15Ω resistor $\text{I}_{15}=\text{I}_{1} .\frac{10}{10+15}=1 \times \frac{2}{5}=0.4 {\text{A}}$ ∴ Voltage drop across15Ω resistor, $\text{V}_{15}=\text{I}_{15} \times 15=0.4 \times 15=6 {\text{V}}$

Given, current, $I=10 {A}$ Cross-sectional area, $A=5 {mm}^{2}=5 \times 10^{-6} {m}^{2}$ Drift velocity, $v_{d}=2 \times 10^{-3} {ms}^{-1}$ The value of current flowing through a conductor can be given by $I=n e A v_{d}$....(i) where, $n=$ number of free electrons and ${e}=$ charge on an electron Putting all the given values in Eq. (i) we get $10=n \times 1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}$ $\Rightarrow \;\; n=\;\frac{10}{1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}}$ $=0.625 \times 10^{28}=625 \times 10^{25}$

From graph voltage at t = 3.2 sec is 6 volt. ${i}=\frac{6-5}{1}$ ${i}=1 {A}$

Given that, initial current, $\text{I}_{1}=4 {\text{A}}$ Final current, $\text{I}_{2}=2 \text{I}_{1}=8 {\text{A}}$ Initial heat dissipated, $\text{H}_{1}=192 {\text{J}}$ Initial time, $t_{1}=1 {\text{s}}$ Final time, $t_{2}=5 {\text{s}}$ Let final heat dissipated = $\text{H}_2$ By Joule's law of heating, $\text{H} \propto \text{I}^2\text{RT}$ Since resistance remains same at initial and final condition, $\therefore \frac{\text{H}_{2}}{\text{H}_{1}}=\frac{\text{I}_{2}^{2} \text{R} t_{2}}{\text{I}_{1}^{2} \text{R} t_{1}}=\frac{\text{I}_{2}^{2} t_{2}}{\text{I}_{1}^{2} t_{1}}$ Substituting the given values, we get $\frac{\text{H}_{2}}{192}=(\frac{8}{4})^{2} \times \frac{5}{1}$ $\Rightarrow \text{H}_2$ = 3840 J

Terminal voltage ${v}={i} {R}=\frac{{ER}}{{R}+{r}}$ $1^{{st}} \to 1.25=\frac{{E}(5)}{5+{r}}$ ........(i) $2^{\text{nd }} \to 1=\frac{E(2)}{2+r} \......$ (ii) By (i) and (ii) ${r}=1 \Omega, {E}=\frac{3}{2} {V}=\frac{15}{10} {\text{volt}}$ $\Rightarrow{x}=15$

Given all resistances have same resistance $R$. Now, we can redraw the circuit as belowLet resistances be $R_{1}, R_{2}, R_{3}$ and $R_{4}$. $\because$ $\;\frac{R_{1}}{R_{3}}=\;\frac{R_{2}}{R_{4}}$ So, circuit will behave as a Wheatstone bridge and no current will flow through midthle resistor. $\therefore \;\; R_{ {eq}} \;\; =\;\frac{(R_{1}+R_{2})(R_{3}+R_{4})}{(R_{1}+R_{2})+(R_{3}+R_{4})}$ $\;\; =\;\frac{(R+R)(R+R)}{(R+R)+(R+R)}$ $\;\; =R$

Max. voltage that can be measured by this potentiometer will be equal to potential drop across ${AB}$ ${R}_{{AB}}=10 \\times 0.1 \\times 100=100 {ohm} .$ ∴ ${V}_{{AB}}=\frac{6}{20+100} \\times 100$ $=6 \\times \frac{100}{120}=5 {V}$

Given, number of divisions, $n=50$Voltage, ${V}=50 {mV}=50 \times 10^{-3} {V}$Current sensitivity, $k=\;\frac{2 div}{ {mA}}=\;\frac{2 div}{10^{-3} {A}}$ $\therefore$ Current, $I=\;\frac{n}{k}=\;\frac{50 \times 10^{-3}}{2}=25 \times 10^{-3} {A}$By using Ohm's law,Resistance $(R)=\;\frac{\;\text{ Potential }\;(V)}{\;\text{ Current }\;(l)}$ $=\;\frac{50 \times 10^{-3}}{25 \times 10^{-3}}=2 Ω$

Given, Heat energy, $H_{1}=500 {J}$ Initial current, $I_{1}=1.5 {A}$, final current, $I_{2}=3 {A}$ and time, $t=20 {s}$ According to Joule's law of heating, $H=I^{2} R t$ $\Rightarrow \;\; H_{1}=l_{1}^{2} R t$.....(i) and $H_{2}=I_{2}^{2} R t$....(ii) [ Resistance and time being the same in both cases] Dividing Eq. (i) by Eq. (ii), we get $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=\;\frac{I_{1}^{2} R t}{I_{2}^{2} R t} \Rightarrow \;\frac{H_{1}}{H_{2}}=(\;\frac{I_{1}}{I_{2}})^{2}$ $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=(\;\frac{1.5}{3})^{2}=(\;\frac{15}{30})^{2} \Rightarrow \;\frac{H_{1}}{H_{2}}=(\;\frac{1}{2})^{2}$ $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=\;\frac{1}{4} \Rightarrow \;\frac{500}{H_{2}}=\;\frac{1}{4} \Rightarrow H_{2}=500 \times 4$ $\Rightarrow \;\; H_{2}=2000 {J}$

$\frac{20-{V}_{0}}{5}+\frac{0-{V}_{0}}{5}+\frac{20-{V}_{0}}{2}=0$ $4+10=\frac{2 {V}_{0}}{5}+\frac{{V}_{0}}{2}$ $14=\frac{4 {V}_{0}+5 {V}_{0}}{10}$ ${V}_{0}=\frac{140}{9} {Volt}$ Potential difference across $2 \Omega$ resistor is $20-{V}_{0}$ That is $(20-\frac{140}{9})$ Volt Hence answer is $(\frac{40}{9})$ Volt

Let the resistance of iron wire be $\text{R}_{1}$ and that of copper nickel alloy wire be $\text{R}_{2}$ $r_{1}=r_{2}=1 {\text{mm}}=10^{-3} {\text{m}}$ $\rho _{1}=12 \\mu \Omega{\text{cm}}$ $=12 \times 10^{-6} \Omega{\text{cm}}$ $=12 \times 10^{-8} \Omega{\text{m}}$ $\rho _{2}=51 \\mu \Omega{\text{cm}}$ $=51 \times 10^{-6} \Omega{\text{cm}}$ $=51 \times 10^{-8} \Omega{\text{m}}$ For parallel combination, $\text{R}_{{\text{eq}}}=\frac{\text{R}_{1} \text{R}_{2}}{\text{R}_{1}+\text{R}_{2}}$ $3={\frac{\rho _{1} \text{l}}{\pi r_{1}^{2}} \frac{\rho _{2} \text{l}}{\pi r_{2}^{2}}}/{\frac{\rho _{1} \text{l}}{\pi r_{1}^{2}}+\frac{\rho _{2}}{\pi r_{2}^{2}}}$ $\Rightarrow 3={\frac{(12 \times 10^{-8}) \text{l}}{\pi \times 10^{-6}}\times \frac{51 \times 10^{-8} \text{l}}{\pi \times 10^{-6}}}/{\frac{12 \times 10^{-8} \text{l}}{\pi \times 10^{-6}}+\frac{51 \times 10^{-8} \text{l}}{\pi \times 10^{-6}}}$ On solving, l = 97 m

Given, specific resistance for wire 1 , $\rho _{1}=6 Ω- {cm}$ Specific resistance for wire 2, $\rho _{2}=3 Ω- {cm}$ Resistance, $R=\;\frac{\rho l}{A}$ For parallel connections, $R_{\;\text{net }\;}=\;\frac{R_{1} R_{2}}{R_{1}+R_{2}}$ $\Rightarrow \;\frac{\rho I}{2 A}=\;{\;\frac{\rho _{1} I}{A} \times \frac{\rho _{2} I}{A}}/{\frac{\rho _{1} I}{A}+\;\frac{\rho _{2} I}{A}}$ $\Rightarrow \;\; \;\frac{\rho }{2}=\;\frac{\rho _{1} \rho _{2}}{\rho _{1}+\rho _{2}}$ $\Rightarrow \;\; \;\frac{\rho }{2}=\;\frac{6 \times 3}{6+3}$ $\rho =4 Ω- {cm}$ Hence, the value of $\rho$ to the nearest integer is 4 .

Initially, the resistance of wire is $R_{1}=\rho L {/} A$ In second case, Length, $I^{'}=2 I$ Area, $A^{'}=\;\frac{A}{2}$ $\therefore \;\; R_{2}=\;\frac{\rho l^{'}}{A^{'}}=\;\frac{\rho (2 l)}{(A {/}2)}=\;\frac{4 \rho l}{A}$ According to Ohm's law, $I=\;\frac{V}{R_{2}}$ $\Rightarrow \;\; l=\;\frac{V}{4 \rho l{/}A}=\;\frac{1}{4} \;\frac{V A}{\rho l}$ This is the required value of resultant current.

{I}={\vec{J}} \. {\vec{A}}={J} {A} \cos (\theta ) $5={J}(\frac{4}{100}) \\times \cos (60)$ ${J}=5 \\times 50=250 {A} / {m}^{2}$ Now, ${\vec{E}}=\rho \.{\vec{J}}$ $=44 \\times 10^{-8} \\times 250=11 \\times 10^{-5} {V} / {m}$

For case I, The potential difference of the uniform wire, V =240 V The resistance of the uniform wire, $\text{R}_1$ = 36 Ω The power dissipation in the first case, $\text{P}_{1}=\frac{\text{V}^{2}}{\text{R}_{1}}=\frac{(240)^{2}}{36}$ For case II, The resistance of each half, $\text{R}_{2}=\frac{\text{R}_{1}}{2}=\frac{36}{2}=18 \Omega$ $\text{P}_{2}=\frac{\text{V}^{2}}{\text{R}_{2}}+\frac{\text{V}^{2}}{\text{R}_{2}}$ $=\frac{(240)^{2}}{18}+\frac{(240)^{2}}{18}=\frac{(240)^{2}}{9}$ Thus, the ratio of the total power dissipation in the first case to thesecond case $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{(240)^{2} / 36}{(240)^{2} / 9}$ \Rightarrow $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{1}{4}$ Comparing with, $\frac{\text{P}_{1}}{\text{P}_{2}}=\frac{1}{x}$ The value of the x = 4.

Given, charge passing through circuit, $q=20 {C}$ Potential difference between two plates, $V=15 {V}$ Let $W$ be the amount of work done by battery. $\therefore \;\; W=q V=20 \times 15=300 {J}$

Current is constant in conductor ${i}=$ constant Resistance of element ${dR}=\frac{\rho {dx}}{\pi {r}^{2}}$ ${dV}={idR}=\frac{{i} \rho {d} {x}}{\pi {r}^{2}}$ ${E}=\frac{{d} {V}}{{d} {x}}=\frac{{i} \rho }{\pi {r}^{2}}$ $\& \ V_{{d}}=\frac{{eE} \tau }{{m}}$ $\therefore {V}_{{d} \ \propto } {E}$ → ${E} \ \propto \frac{1}{{r}^{2}}$ if ${r}$ decreases, ${E}$ will increase $\therefore {V}_{{d}}$ will increase

here assume current as By KVL in outer loop $9-12 {i}-4 {i}=0$ $16 {i}=9$ $8 {i}=\frac{9}{2}=4.5$ $=45 \\times 10^{-1}$