Science Answer the questions.[Phy-3M]

1. When either coil is used separately, the circuit $\text{(I)}= \frac{\text{V}}{\text{R}} = \frac{\text{220V}}{24\Omega} = 9.2 \text{A.}$

2. When two coils are used in series total resistance 

$ \text{(R)}=\text{R}_1 + \text{R}_2 = 24 + 24 =48 \Omega$

Current flowing $\text{(I)} = \frac{\text{V}}{\text{R}} = \frac{220 \text{V}}{ 48\Omega} = 4.6 \text{A.}$

3. When two coils are joined in parallel. Total resistance $\text{(R)}=\frac{1}{24} + \frac{1}{24}$

$= \frac{2}{24},\text{ R} =12 \Omega$

Current $\text{(I)} = \frac{\text{V}}{\text{R}} = \frac{220 \text{V}}{12} = 18.3\text{A.}$

$\text{R}_1=\frac{(220)^2}{10}=4840 \ \Omega$

$=\frac{220}{5}=44 \ \Omega$

1. One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to another.
2. Work = charge x potential difference $(\text{OR }\text{W = Q} \times \text{V})$

$V = \frac{W}{Q}$

$V = \frac{\text{100 J}}{\text{20 C}}$

$\therefore V = \text{5 volt}$

$\text{I} = \frac{\text{P}}{\text{V}}$

$\text{I}_{1} = \frac{\text{60 W}}{\text{220 V}} = \frac{3}{11}\text{A}$

$\text{I}_{2} = \frac{\text{40 W}}{\text{220 V}} = \frac{2}{11}\text{A}$

$\text{I} = \text{I}_{1} + \text{I}_{2} = \frac{3}{11} + \frac{2}{11} = \frac{5}{11} \text{A} = 0.45 \text{ A}$

$\text{E} = \text{P} \times \text{t}$

$= \text{(40 W + 60 W)} \times 1\text{ h} = \text{100 Wh or 0.1 kWh}$

1. In series

2. In parallel arrangement

2. For series arrangement:

$\text{R = R}_{1} + \text{R}_{2} = \text{5 }\Omega + \text{10 }\Omega = \text{15 }\Omega$

$\text{I} = \frac{\text{V}}{\text{R}} = \frac{\text{6V}}{15\Omega} = 0.4\text{A}$

For parallel arrangement:

$\frac{1}{\text{R}} = \frac{1}{\text{R}_{1}} + \frac{1}{\text{R}_{2}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$\therefore \text{R} = \frac{10}{3}\Omega$

$\text{I} = \frac{\text{V}}{\text{R}} = \frac{6\text{V}}{\frac{10}{3}\Omega} = 1.8 \text{ A}$

$\text{I} = \frac{\text{V}}{\text{R}}$

$\text{I}_{1} = \frac{\text{6V}}{5\Omega} = 1.2\text{A}$

$\text{I}_{2} = \frac{6\text{V}}{10\Omega} = 0.6\text{A}$

$\text{I}_{3} = \frac{6\text{V}}{30\Omega} = 0.2\text{A}$

$\text{I} = \text{I}_{1} + \text{I}_{2} + \text{I}_{3} = 1.2\text{A} + 0.6\text{A} + 0.2\text{A} = \text{2A}$

$\frac{1}{\text{R}} = \frac{1}{\text{R}_{1}} + \frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}$

$=\frac{1}{5} + \frac{1}{10} + \frac{1}{10}+\frac{1}{30} = \frac{10}{30} = \frac{1}{3}$

$\therefore \text{R} = 3\Omega$

Alternate Answer

$\text{V} = \text{I R}$

$\therefore \text{R} = \frac{\text{V}}{\text{I}} = \frac{6\text{V}}{\text{2A}} = 3\Omega$

1. According to Joule’s law of heating, heat produced in a wire is directly proportional to,

1. Square of current (I²)
2. Resistance of wire (R)
3. Time (t), for which current is passed

Thus,

H = I² × R × t … (Joule’s law of heating)

2. Current drawn from the first lamp is given by,

$\text{P}_1=\text{V}\times\text{I}_1$

$\therefore\ \text{I}_1=\frac{\text{P}_1}{\text{V}}=\frac{100}{220}=0.45\text{A}$

Current drawn from the second lamp is given by,

$\text{P}_2=\text{V}\times\text{I}_2$

$\therefore\ \text{I}_2=\frac{\text{P}_2}{\text{V}}=\frac{60}{220}=0.27\text{A}$

Thus the total current drawn by two lamps from the line,

which are connected in parallel to each other for supply voltage 220V is,

$\text{I}=\text{I}_1+\text{I}_2=0.45+0.27=0.72\text{A}$

$9\Omega$ are connected in the circuit.

1. When one resistor is connected in series with the other two resistors which are connected in parallel to each other, the equivalent resistance in the circuit is,

$\frac{1}{\text{R}_\text{p}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}=\frac{1}{9}+\frac{1}{9}=\frac{2}{9}=0.22$

$\Rightarrow\ \text{R}_\text{p}=4.5\Omega$

$\text{R}_\text{s}=\text{R}_\text{p}+\text{R}_3=4.5\Omega+9\Omega=13.5\Omega$

$\therefore\ \text{R}_{\text{eq}}=13.5\Omega$

Thus, by connecting the resistors in this combination, the equivalent resistance in the circuit, $\text{R}_{\text{eq}}=13.5\Omega.$

2. When one resistor is connected in parallel with the other two resistors which are connected in series with each other, the equivalent resistance in the circuit is,

$\text{R}_{\text{s}}=\text{R}_1+\text{R}_2=9\Omega+9\Omega=18\Omega$

$\frac{1}{\text{R}_\text{p}}=\frac{1}{\text{R}_\text{s}}+\frac{1}{\text{R}_3}=\frac{1}{18}+\frac{1}{9}=0.166$

$\Rightarrow\ \text{R}_\text{p}=6\Omega$

$\therefore\ \text{R}_{\text{eq}}=6\Omega$

Thus, by connecting the resistors in this combination, the equivalent resistance in the circuit, $\text{R}_{\text{eq}}=6\Omega.$​​​​​​​

1. $2\text{Cu}_2\text{S}+3\text{O}_2\xrightarrow{\ \ \ \text{Heat}\ \ \ \ }2\text{Cu}_2\text{O}+2\text{SO}_2$

2. $2\text{Cu}_2\text{O}+\text{Cu}_2\text{S}\xrightarrow{\ \ \ \text{Heat}\ \ \ \ }6\text{Cu}+\text{SO}_2$

3. $\text{At anode}\xrightarrow{\ \ \ }\text{Cu}\xrightarrow{\ \ \ }\text{Cu}^{2+}+2\bar{\text{e}}$

$\text{At Cathode}\xrightarrow{\ \ \ }\text{Cu}^{2+}+\text{2e}\xrightarrow{\ \ \ }\text{Cu}$

1. The mathematical expression of the Joules Law of heating is:

H = I² Rt

Here, H is a heating effect, I is the current flowing through the device, and t is the time taken

2. Given,

Amount of charge transferred = 96000C

Time taken = 2hrs = 2 × 60 × 60sec

= 7200sec

Potential difference = 40V

Heat generated = V × i × t

And we know that;

$\text{i}=\frac{\text{Q}}{\text{t}}$

So, H = VQ

= 40 × 96000

= 3.84 × 10⁶J.

$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$

$\frac{1}{\text{f}}=\frac{1}{(-50)}-\frac{1}{(-25)}=\frac1{50}$

$\text{f}=50\text{cm or }0.50\text{m}$

$\text{p}=\frac{1}{\text{f}}=\frac{1}{0.5}=+2\text{D}.$

1. $\text{V}\alpha\text{ I}\text{ or }\frac{\text{V}}{\text{I}}=\text{Constant}$

$\text{Or}\text{ V}=1\text{R}$ 

2. given I = 0.35A,

V = 1.4V

$\text{R}=\frac{\text{V}}{\text{I}}$

$=\frac{1.4}{0.35}$

$=4.'\Omega$

$\text{R}=\frac{\text{V}}{\text{I}}$

$=\frac{0.4}{0.1}=4\Omega$

$4\Omega.$

Nichrome wire follows ohm's law thus it behaves as an ohmic conductor.

The circuit diagram corresponding to the above graph is as shown below.

$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$

$=2.8\times10^{-8}\times\frac{2}{1.55\times10^{-6}}$

$0.036\Omega$

$\frac{1}{\text{R}}=\frac{1}{30}+\frac{1}{50}$

$\frac{1}{\text{R}}=\frac{80}{1500}$

$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24 }$

1. To obtain highest resistance, all the four resistances must be connected in series arrangement. In that case resultant

$\text{R} = \text{R}_1 + \text{R}_2 + \text{R}_3 $

$= 4 + 8 + 12 + 24 =48 \Omega$

2. To obtain lowest resistance, all the four resistance must be connected in parallel arrangement.

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}+\frac{1}{\text{R}_3}$

$\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{12}{24} \Omega$

$=\frac{24}{12}=2 \Omega$

$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

$\frac{0.06}{3600}\text{kwh}$

$=1.67\times\text{e}-5\text{kwh}.$

$\text{R}_1=2\Omega,\text{R}_2=3\Omega$

$\text{I}=\frac{\text{V}}{\text{R}_\text{tot}}=\frac{10}{5}=2\text{A}$

$\frac{1}{0.1} = \text{R}$

$\frac{2.2}{0.2} =\text{R}$

$\frac{3}{0.6} = \text{R}$

$\text{p}.\text{d}.=\frac{\text{Work done}}{\text{Charge moved}}=\frac{200}{25}=8\text{V}$

$\text{I}=\frac{\text{Q}}{\text{t}}=\frac{25}{10}=2.5\text{A}$

$\text{I}=\frac{\text{Q}}{\text{T}}$

$\Rightarrow1\text{A}=\frac{\text{Q}}{1\text{s}}$

$\Rightarrow\text{Q}=1\text{C}$

$\text{I} = \frac{\text{Q}}{\text{T}}$

$\frac{1}{\text{R'}_\text{p}}=\frac{1}{\text{R}_\text{AC}}+\frac{1}{\text{R}_\text{ED}}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$

$\Rightarrow\text{R'}_\text{p}=15\Omega$

$\text{R'}_\text{s}=\text{R'}_\text{p}+\text{R'}_\text{BC}=15+15=30\Omega$

$\frac{1}{\text{R''}_\text{p}}=\frac{1}{\text{R}_\text{AB}}+\frac{1}{\text{R'}_\text{s}}$

$=\frac{1}{15}+\frac{1}{30}=\frac{1}{10}$

$\therefore \text{R''}_\text{p}=10\Omega$

$\text{I}=\frac{\text{V}}{\text{R''}_\text{P}}=\frac{3}{10}=0.3\text{A}$

$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\text{ or R}=2\Omega$

Therefore, the above circuit can be redrawn as given below:

1. $\text{Conductors } — 10^{-6} \text{ to } 10^{-8}\Omega$
2. $\text{Alloys } — 10^{-6} \Omega$
3. $\text{Insulators} — 10 ^{12} \text{ to } 10^{17}Ωm. $