Current Electricity Practice Questions

Given, radius of cylindrical wire, $r=0.5 {mm}=0.5 \times 10^{-3} {m}$ Conductivity, $\sigma =5 \times 10^{7} {S} / {m}$ Electric field, $E=10 {mV} / {m}=10 \times 10^{-3} {V} / {m}$ We know that current density, $\therefore J \;\; =\sigma E$ $\;\; =5 \times 10^{7} \times 10 \times 10^{-3}=5 \times 10^{5} {A} / {m}^{2}$ Also, $J=I / {A} \Rightarrow I=J {A}$ $\Rightarrow$ $I \;\; =5 \times 10^{5} \times \pi \times (0.5 \times 10^{-3})^{2}$ $\;\; =5 \times 10^{5} \times \pi \times 25 \times 10^{-8}=125 \pi \times 10^{-3}$ $\Rightarrow \;\; x^{3} \pi {mA}=125 \pi {mA} \Rightarrow x^{3}=5^{3}$ $\Rightarrow$ $x=5$

Given, number of cells, n = 5 Internal resistance of each cell, r = 1Ω Emf of each cell, e = 5 V When all cells are connected in parallel as shown below. Potential will remain same as, $\text{V}_{\text{P}}=5 {\text{V}}$ Net resistance in parallel combination will be given as $\text{R}_{\text{p}}=\text{R}+[\frac{1}{\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}}]=\text{R}+\frac{r}{5}$ $\text{R}_{\text{p}}=\text{R}+\frac{1}{5}$ (\because r = 1 Ω) ...(i) When all cells are connected in series as shown below The net potential will increase as cells are connected in series, $\text{V}_{\text{S}}=5+5+5+5+5=25 {\text{V}}$ The net resistance of circuit will be equivalent of sum of all resistances as all are connected in series. $\text{R}_{\text{s}}=r+r+r+r+r+\text{R}=5 r+\text{R}$ $\text{R}_{\text{s}}=5+\text{R}$ ( \because r = 1Ω) ...(ii) By Ohm's law, current flowing in circuit is given as $\text{I}=\text{V} / \text{R}$ As current in both series and parallel combination is same, ${\text{V}_{\text{p}}}/{\text{R}_{\text{p}}}={\text{V}_{\text{s}}}/{\text{R}_{\text{s}}}$ $\Rightarrow \frac{5}{\text{R}+\frac{1}{5}}=\frac{25}{5+\text{R}}$ [From Eqs. (i) and (ii)] $\Rightarrow 25+5 \text{R}=25 \text{R}+5$ ⇒ R = 1 Ω

Figure of Merit $= {C}=\frac{{i}}{\theta }$ $= {C}= \frac{6 \\times 10^{-3}}{2}$ $=3 \\times 10^{-3} {Am}^{2}$

${i}_{1}= \frac{40}{40+60}=0.4$ $\ {i}_{2}= \frac{40}{90+110}= \frac{1}{5}$ $\ {v}_{\ {B}}+ {i}_{1}(40)- {i}_{2}(90)= {v}_{ {D}}$ $\ {v}_{ {B}}- {v}_{ {D}}= \frac{1}{5}(90)- \frac{4}{10} \\times 40$ ${v}_{ {B}}- {v}_{ {D}}=18-16=2$

220 I = P = 15 × 45 + 15 × 100 + 15 × 10 + 2 × $10^3$ $I=4325/220=19.66$ $I≃ 20 A$

In balancing$R/S=25/75$New resistance $R'={ \rho l}/A$ $=\frac{\rho \times l/2}{A/4}={\rho l}/2\times 4A$R' = 2R${2R}/S=l'/{100-l'}$ $2\times 1/3=l'/{100-l'}$ $=3l'=200-2l'$ $5l'=200$ $l'=40$

$\ {vi}=10^{3}$ $\ {i}=\ \frac{1000}{220}$ loss $= {i}^{2} {R}=(\frac{50}{11} )^{2} \\times 2$ efficiency $= \frac{1000}{1000+ {i}^{2} {R}} \\times 100=96 \%$

${E}_{{eq}}=\frac{20 \times 10}{17}=\frac{200}{17}$ and ${R}_{{eq}}=\frac{7 \times 10}{17}=\frac{70}{17}$ ${I}=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 {\text{A}}$ from +ve to -ve terminal

Voltage across ${AC}=8 {V}$ ${R}_{ {AC}}=4+4=8 \Omega$ ${i}_{1}={{V}}/{\ {R}_{ {AC}}}= \frac{8}{8}=1 \ {A}$

${P}_{ {R}}=0.5 {W}$ $\Rightarrow {i}^{2} {R}=0.5 {W}$ Also, ${V}= {E}- {ir}$ $2.5=3- {ir}$ $\Rightarrow$ ir $=0.5$ Power dissipated across 'r' : ${P}_{ {r}}= {i}^{2} {r}$ Now iR $=2.5$ $\ {ir}=0.5$ On dividing $: \frac{{R}}{ {r}}=5$ Now $\frac{P_{R}}{P_{r}}= \frac{i^{2} R}{i^{2} r}\Rightarrow \frac{P_{R}}{P_{r}}=\ \frac{R}{r}$ $\Rightarrow \frac{P_{R}}{P_{r}}=5$ $\Rightarrow P_{ {r}}={{P}_{ {R}}}/{5}$ $\Rightarrow {P}_{ {r}}= \frac{0.50}{5}$ $\Rightarrow {P}_{ {r}}=0.10 {W}$ option (4) is correct.

$E_1=E-ir$ $=10-i20=0$ $\Rightarrow i=0.5A$ $E_2 = E - ir$ $=10-0.5\times 5=7.5V$ $E_\text{net} = E_1+ E_2= 7.5V$ $i = i_1+ i_2$ $0.5=7.5/x+7.5/30$ $x = 30 \Omega$

$5 = \lambda l$ where λ is potential gradient & L is total length of wire$5={\DeltaV}/Ll$ $\Delta={5\times L}/l=5\times 12/10=6\text{V}$ = 60 mA × RR = 100 Ω

Let the resistance to be connected is R.For balanced Wheatstone bridge$15\times 4-12\times \frac{10R}{10+R}$ $R=10\Omega$

Let us asssume the potential at ${A}={V}_{ {A}}=0$ Now at junction C, According to KCL ${i}_{1}+ {i}_{3}= {i}_{2}$ $1 { A}+ {i}_{3}=2 { A}$ ${i}_{3}=2 { A}$ Now Analyse potential along ACDB $\ {v}_{ {A}}+1+ {i}_{3}(2)-2= {v}_{ {B}}$ $0+1+2(1)-2= {v}_{ {B}}$ ${v}_{ {B}}=3-2$ ${v}_{ {B}}=1$ Amp

Equivalent resistance of upper branch of circuit R = 2.5 ΩVoltage across upper branch = 1 V$\Rightarrow i=1/2.5=.4A$ $\Rightarrow I_1=0.2A$

$\Rightarrow 1=\ {i}_{ {g}}({G}+ {R}_{1})$ ....(1) $\Rightarrow2=\ {i}_{ {g}}({R}_{1}+ {R}_{2}+ {G} )$ ....(2) (1) % (2) $\Rightarrow \frac{1}{2}=\ \frac{G+R_{1}}{G+R_{1}+R_{2}}$ $G+R_{1}+R_{2}=2 G+2 h_{1}$ $\ (R_{2}=G+R_{1} )$

Ammeter :- In series connection, the same current flows through all the components. It aims at measuring the current flowing through the circuit and hence, it is connected in series. Voltmeter :- A voltmeter measures voltage change between two points in a circuit, So we have to place the voltmeter in parallel with the circuit component.