Current Electricity Practice Questions

when switch ${S}_{1}$ and ${S}_{2}$ are closed$\frac{12 \\times 6}{12+6}+2+\frac{6 \\times 12}{6+12}$ $\frac{72}{18}+2+\frac{72}{18}=4+2+4=10 \Omega$

${R}=\frac{{R}_{1} {R}_{2}}{{R}_{1}+{R}_{2}}=\frac{l}{{A}} \.\frac{\rho _{1} \rho _{2}}{\rho _{1}+\rho _{2}}$ ${R}=\frac{25 \\times 10^{-2}}{3 \\times 10^{-6}} \\times \frac{1.7 \\times 2.6 \\times 10^{-16}}{4.3 \\times 10^{-8}}$ ${R}=0.858 {m} \Omega$

Given, length of $A B=10 {m}=1000 {cm}$ Length of one ${arm}=\;\frac{1000}{10}=100 {cm}$ For no deflection, In first case, $l_{1}=3 \times 100+80=380 {cm}$ In 2 nd case, $I_{2}=7 \times 100+60=760 {cm}$ As we know that in balanced potentiometer, $\Rightarrow \;\; \;\frac{a}{b}=\;\frac{380}{760}=\;\frac{1}{2}$ $\therefore \;\; a=1$

Given, power rating of bulb, $\text{P}_{\text{B}}=500 {\text{W}}$ Voltage across bulb, $\text{V}_{\text{B}}=100 {\text{V}}$ Supply voltage, $\text{V}_{\text{S}}=200 {\text{V}}$ If a resistance R is attached in series with the bulb, then the voltage across resistance will be 100 V. Now, current flowing in circuit when bulb delivers power of 500 W is given as $\text{P}_{\text{B}}=\text{V}_{\text{B}} \text{I}$ $\Rightarrow 500=100 \times \text{I}$ $\Rightarrow \text{I}=5 {\text{A}}$ Same amount of current will flow from the resistance as it is connected in series. Using Ohm's law, V = IR ⇒ 100 = 5 × R ⇒ R = 20Ω Thus, the resistance connected in series is 20Ω.

The given circuit can be drawn as Since in series combination, the current through each resistance remains same. So, equivalent resistance of the circuit is given as $R_{\;\text{equivalent }\;}=R+r_{1}+r_{2}$ and equivalent emf, $E_{\;\text{equivalent }\;}=2 E+E=3 E$ From Ohm's law, I = {E_{\;\text{equivalent}}/{R_{\;\text{equivalent}\; }}\Rightarrow I=\frac{3 E}{R+r_{1}+r_{2}} When potential difference is zero across the first cell, then potential positive terminal is equal to the potential at negative terminal. $V_{P}=V_{N}$ $2 E=I r_{1}$ Substituting the values in the above equation, we get $2 E=\;\frac{3 E}{R+r_{1}+r_{2}} r_{1}$ $2 R+2 r_{1}+2 r_{2}=3 r_{1} \Rightarrow R=\;\frac{r_{1}-2 r_{2}}{2}$ $\Rightarrow \;\; R=\;\frac{r_{1}}{2}-r_{2}$

The circuit can be shown as below The current through the circuit, $I=\;\frac{\epsilon _{1}-\epsilon _{2}}{r_{1}+r_{2}}=\;\frac{6-4}{10}=\;\frac{1}{5} {A}$ $\therefore$ Potential difference across points $X$ and $Y$ is $V_{X X}=E_{2}+I r_{2}=4+\;\frac{1}{5} \times 8=5.6 {V}$

Since, $\lambda v=c=$ constant where, $\lambda =$ wavelength of light and $v=$ frequency of light. Red light and blue light have different wavelengths and different frequencies but same speed. $\Rightarrow \;\; R=\;\frac{H}{I^{2} T}$ .....(i) Substituting the given values in Eq. (i), we get $R=\;\frac{10 \times 10^{-3}}{(2 \times 10^{-3})^{2} \times 1}$ $\Rightarrow \;\; R=\;\frac{10^{-2}}{4 \times 10^{-6}} \Rightarrow R=0.25 \times 10^{4}$ $\Rightarrow \;\; R=2500 Ω$

The given circuit diagram can be drawn as Redrawing the above circuit diagram as Equivalent resistance of the circuit between point A and B is given as ${1}/{\text{R}_{{\text{eq}}}}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}+\frac{1}{\text{R}_{4}}$ $=\frac{1}{4}+\frac{1}{12}+\frac{1}{6}+\frac{1}{8}$ $=\frac{6+2+4+3}{24}$ $=\frac{15}{24}$ $=\frac{24}{15} \Omega=1.6 \Omega$ $\text{R}_{{\text{total} }}=1.6+0.6=2.2 \Omega$ $\text{P}={\text{V}^{2}}/{\text{R}_{{\text{total} }}}=\frac{2.2 \times 2.2}{2.2}$ = 2.2 W

Given, resistance of wire, $R=1 Ω$ Increase in length, $\Delta l=25 \%$ Let $I_{1}, I_{2} A_{1}, A_{2}, V_{1}$ and $V_{2}$ be the length, area and volume, respectively. As, volume remains same and $V=A l$ $\therefore \;\; A_{1} l_{1}=A_{2} l_{2}$ $\Rightarrow \;\; A_{2}=\;\frac{A_{1} I_{1}}{I_{2}}$ a nd resistance, $R=\;\frac{\rho l}{A}$ $\therefore \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{\rho I_{2}}{A_{2}} \times \;\frac{A_{1}}{\rho I_{1}}$ $\Rightarrow \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{I_{2} A_{1}}{A_{2} l_{1}}=\;\frac{1.25 I_{1} \cdot A_{1} \times I_{2}}{A_{1} l_{1} \times I_{1}}( \because I_{2}=1.25 I_{1})$ $\Rightarrow \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{(1.25 I_{1})^{2} A_{1}}{I_{1}^{2} A_{1}} \Rightarrow R_{2}=(1.25)^{2} R_{1}$ $\;\text{ Now, }\; \;\frac{\Delta R}{R} \times 100 \;\; =\;\frac{R_{2}-R_{1}}{R_{1}} \times 100$ $\;\; =\;\frac{(1.25)^{2} R_{1}-R_{1}}{R_{1}} \times 100$ $\therefore \;\; \;\frac{\Delta R}{R} \times 100 \;\; =\;\frac{(1.25)^{2}-1}{1} \times 100$ $\;\; =56.25 \% ∼eq 56 \%$

Let two resistors have resistances $R_{1}$ and $R_{2}$, respectively. As per question, equivalent resistance of series combination is $s$ $\Rightarrow$ $s=R_{1}+R_{2}$.....(i) and equivalent resistance of parallel combination is $p$ $\Rightarrow \;\; p=\;\frac{R_{1} R_{2}}{R_{1}+R_{2}}$....(ii) According to the question, $s=n p$....(iii) From Eqs. (i),(ii) and (iii), we get $R_{1}+R_{2}=n \;\frac{R_{1} R_{2}}{(R_{1}+R_{2})} \Rightarrow n(R_{1} R_{2})=(R_{1}+R_{2})^{2}$ $\Rightarrow n=\;\frac{(R_{1}+R_{2})^{2}}{R_{1} R_{2}}$ For $n$ to be minimum, $R_{1}=R_{2}=R$ $\Rightarrow \;\; n=\;\frac{(R+R)^{2}}{R \cdot R}=\;\frac{(2 R)^{2}}{R^{2}}=\;\frac{4 R^{2}}{R^{2}} \Rightarrow n=4$

Power, ${P}={{V}^{2}}/{{R}_{{B}}}$ ${R}_{{B}}=\frac{{V}^{2}}{{P}}=\frac{100 \\times 100}{200}$ ${R}_{{B}}=50 \Omega$ To produce same power, same voltage (i.e. $100 {V}$ ) should be across the bulb Hence, ${R}={R}_{{B}}$ ${R}=50 \Omega$

Given, $\text{R}_{1}=(4 \±0.8) \Omega$ $\text{R}_{2}=(4 \± 0.4) \Omega$ Equivalent resistance when the resistors are connected in parallel isgiven by ${1}/{\text{R}_{{\text{eq}}}}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}$ \Rightarrow ${1}/{\text{R}_{\text{eq}}}=\frac{1}{4}+\frac{1}{4}$ $\text{R}_{{\text{eq}}}=2 \Omega$ Now, ${\Delta \text{R}_{{\text{eq}}}}/{\text{R}_{{\text{eq}}}^{2}}$ $=\frac{\Delta \text{R}_{1}}{\text{R}_{1}^{2}}+\frac{\Delta \text{R}_{2}}{\text{R}_{2}^{2}}$ Substituting the values in the above equation, we get ${\Delta \text{R}_{{\text{eq}}}}/{4}=\frac{0.8}{16}+\frac{0.4}{16}$ $\Delta \text{R}_{\text{eq}}=0.3 \Omega$ ∴ The equivalent resistance in parallel combination is $\text{R}_{{\text{eq}}}=(2 \±0.3) \Omega$.

Given, the length of the water pipe, L = 1 m The cross-sectional area of the water pipe, A$=1 {\text{cm}}^{2}=10^{-4} {\text{m}}^{2}$ The temperature of the ice = -10ºC Current passing in the conductor, I = 0.5 A Resistance of the conductor, R = 4 kΩ The latent heat of fusion for ice, $,\text{L}_{\text{i}}=3.33 \times 10^{5} {\text{J}} / {\text{kg}}$ The density of the ice, d $=1000 {\text{kg}} / {\text{m}}^{3}$ The specific heat of the ice, $\text{c}_{\text{p}, { \text{ice} }}=2 \times 10^{3} {\text{J}} / {\text{kg}}$ Heat required to melt the ice at 10°C to 0°C $\text{Q}=\text{m} \text{c}_{\text{p}} \Delta \text{T}+\text{m} \text{L}_{\text{i}}$ $\text{Q}=\text{d} \text{V} \text{c}_{\text{p}} \Delta \text{T}+\text{d} \text{V} \text{L}_{\text{f}}$ $=1000 \times 10^{-4} \times 2 \times 10^{3} \times (10) +1000 \times 10^{-4} \times 3.33 \times 10^{5}$ $(\because \text{V}=\text{A} \times \text{L})$ = 35300 J According to the Joule's law of heating, $\text{H}=\text{I}^{2} \text{R} \text{t}$ \Rightarrow $35300=(0.5)^{2}(4000)(\text{t})$ t = 35.3 s Thus, the minimum time required to melt the ice is 35.3 s.

Given, Shunt resistance, $\text{R}_{{\text{sn}}}=5 \Omega$ Let galvanometer resistance = G Total, current = I Voltage across galvanometer $(\text{V}_{\text{G}})=$ Voltage across shunt resistance $\text{R}_{{\text{sn} }}$ $\text{I}_{\text{G}} \times \text{G}=\text{I}_{{\text{sn}}} \times \text{R}_{{\text{sn}}}$ $0.021 \times \text{G}=0.981 \times 5$ $\Rightarrow \text{G}=\frac{98 \times 5}{2}=49 \times 5=245 \Omega$

${E}_{1}={k} l_{1}$ .........(i) ${E}_{1}+{E}_{2}={k} l_{2}$ ........(ii) $\frac{{E}_{1}}{{E}_{1}+{E}_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}$ $8 {E}_{1}=5 {E}_{1}+5 {E}_{2}$ $3 {E}_{1}=5 {E}_{2}$ $\frac{E_{1}}{E_{2}}=\frac{5}{3}$

Electrical circuit is shown in the diagram. Now, let's draw the equivalent circuit.Equivalent resistance of the circuit, $R_{ {eq}}=\;\frac{50 \times 20}{50+20}$ $R_{ {eq}}=\;\frac{100}{7} Ω$ According to voltage division rule, voltage across the $10 Ω$ resistance of the circuit, $V_{10 Ω}=170 \times (\;\frac{10}{10+\;\frac{100}{7}})$ $V_{10 Ω}=70 {V}$ The value of the $x$ to the nearest integer is 70 .

Given, the potential difference applied across the ends of the conductor, $V=5 {V}$ The length of the conductor, $L=10 {cm}$ The measured value of the current in the conductor, $I=2 {A}$ The diameter of the conductor, $d=5 {mm}$ As we know that, $R=\;\frac{\rho l}{A}$ Using Ohm's law, $V=I R \Rightarrow R=\;\frac{V}{I}$ $\;\frac{\rho l}{A}=\;\frac{V}{l} \Rightarrow \rho =\;\frac{V}{l l}(\;\frac{\pi d^{2}}{4})$ In error form, $\;\frac{\Delta \rho }{\rho }=\;\frac{\Delta V}{V}+\;\frac{\Delta l}{l}+\;\frac{\Delta l}{l}+2 \;\frac{\Delta d}{d}$ $\Rightarrow \;\; \;\frac{\Delta \rho }{\rho }=\;\frac{0.1}{5}+\;\frac{0.01}{2}+\;\frac{0.1}{10}+2 \;\frac{(0.01)}{(5)} \Rightarrow \;\frac{\Delta \rho }{\rho }=0.039$ $\;\frac{\Delta \rho }{\rho } \times 100=3.9 \%$ The maximum permissible percentage error in the resistivity of the conductor is $3.9 \%$.

As, A is directly connected to the positive terminal of the battery, $V_{A}=10 V$ and $V_{C}=0$ By nodal analysis at $B$, $\;\frac{V_{B}-10}{100}+\;\frac{V_{B}-V_{D}}{15}+\;\frac{V_{B}-0}{10}=0$ $53 V_{B}-20 V_{D}=30$...(i) By nodal analysis at $D$, $\;\frac{V_{D}-10}{60}+\;\frac{V_{D}-V_{B}}{15}+\;\frac{V_{D}-0}{5}=0$ $-4 V_{B}+17 V_{D}=10$....(ii) Solving Eqs. (i) and (ii) by substitution method, we get $V_{D}=0.792 {V} \Rightarrow V_{B}=0.865 {V}$ The current through the galvanometer, $I=\;\frac{V_{B}-V_{D}}{R}$ Substituting the values in the above equation, we get $I=\;\frac{0.865-0.792}{15} \Rightarrow I=4.87 {mA}$