Given, resistance, \(R=2 \Omega\)
Maximum power, \(P_{\max }=18 \mathrm{~W}\)
Maximum current, \(I_{\max }=\) ?
As we know, \( P=I^2 R\)
\(\Rightarrow \quad I=\sqrt{\frac{P}{R}}=\sqrt{\frac{18}{2}}=3 \mathrm{~A}=I_{\max }\)
Maximum current that can flow through \(2 \Omega\) resistor is 3 A . This current divides along B and C because they are in parallel combination.
Voltage across B and C remain same and hence \(I \propto \frac{1}{R}\). Since, B and C have same resistance same current flows through them.
i.e. \(\frac{3}{2}=15, A\) flows through B and C.

(i) Given, power of generator, \(P \approx 2 \mathrm{~kW}\)
Time, t=3h
therefore Energy consumed, \(E=P \times t=2 \times 3=6 \mathrm{kWh}\)
Since, cost of 1 kWh of electrical energy is ₹ 3.50.
Therefore Cost of 6 kWh of electrical energy =6 x 350 =₹ 21.0
(ii) As we know that, Heat, \(H=t^2 R t\)
Here, I is constant, so to reduce the boiling time ( f ), R should be increased.
Since, \(R=\rho \frac{l}{A}\) or \(R \propto l\)
So, the length of the heating element should be increased.

As we know, \(H=I^2 R t\)
or \(H=P \times t \quad\left(\because P=I^2 R\right)\)
From above equation, we conclude that if power is minimum, then heat produced will be minimum.
So, in series combination, we get minimum power \(\left(\frac{V^2}{R}\right)\) and minimum heat. Produced in the circuit.

Here, resistances $R_3$ and $R_4$ are in series, so their equivalent resistance is
\(R  =R_3+R_4 =6+4=10 \Omega\)
The circuit is reduced to

Now, resistances R2 and R' are in parallel, so their equivalent resistance,

\(R^{\prime}=\frac{R_2 R^{\prime}}{R_2+R^{\prime}}=\frac{10 \times 10}{10+10}=\frac{100}{20}=5 \Omega\)
The circuit now becomes

The resistances R₁ and R' are in series, so their equivalent or total resistance of circuit, \(R_{\mathrm{eq}}=R_1+R^{\prime}=7+5=12 \Omega\)
The final circuit is as shown
By Ohm's law, \(I=\frac{V}{R_{\text {eq }}}=\frac{24}{12}=2 \mathrm{~A}\)

(i) 
(ii) Total resistance,
\(\Rightarrow  \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
\(\Rightarrow  \frac{1}{R_T}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30} \)
\(\Rightarrow  R_T=5.45 \Omega\)
(iii) Current, I=V / R
\(I  =\frac{10}{R_T}=\frac{10}{5.45} =1.83 \mathrm{~A}\)

Circuit diagram consisting of battery, resistors and key is

Total voltage of battery, \(V=1.5 \times 4=6.0 \mathrm{~V}\)
Equivalent resistance, \(R=2 \Omega+6 \Omega+16 \Omega=24 \Omega\)
(a) Current in the circuit (ammeter reading)
\(\Rightarrow\) using ohm's law, V=I R
\(I=\frac{V}{R}=\frac{6}{24}=0.25 \mathrm{~A}\)
(b) Voltage across the \(16 \Omega\) resistor (voltmeter reading when key is closed) in series, current is same in all resistors.
Potential across \(16 \Omega\) resistor will be, \(V=I R=0.25 \times 16=4 \mathrm{~V}\)

Resistance of combination of three bulbs in parallel,
\(R_{\mathrm{eq}}=\frac{V}{I}=\frac{4.5}{3}=1.5 \Omega\)
If R is the resistance of each wire, then
\(\frac{1}{R_{e q}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\)or \(\frac{1}{R_{\mathrm{eq}}}=\frac{3}{R}\)
or \(R=3 R_{\mathrm{eq}}=3 \times 15=4.5 \Omega\)
Current in each bulb, \(I=\frac{V}{R}=\frac{4.5 \mathrm{~V}}{4.5 \Omega}=1 \mathrm{~A}\)
(i) When bulb B₁ gets fused, then the currents in B₂ and B₃ remain same \(I_2=I_3=1 \mathrm{~A}\), so their glow remains unaffected.
(ii) When bulb B₂ gets fused, then the current in B₂ becomes zero and currents in B₁ and B₃ remain 1 A.
Total current,
\(I =I_1+I_2+I_3 =1+0+1=2 \mathrm{~A}\)
Current in ammeter \(A_1, I_1=1 \mathrm{~A}\)
Current in ammeter \(A_2, l_2=0\)
Current in ammeter \(\mathrm{A}_3, I_3=1 \mathrm{~A}\)
Current in ammeter \(A, I=2 \mathrm{~A}\)
(iii) When all the three bulbs are connected, then power dissipated,
\(P  =\frac{V^2}{R_{\mathrm{eq}}}=\frac{(4.5)^2}{1.5}  =135 \mathrm{~W}\)

(i) Electric energy:
1. The energy is generated by the movement of electrons from one point to another.
2. The joule or watt-second is the fundamental unit of electrical energy.
3. A joule is defined as one ampere of current flowing through a circuit for one second when a potential difference of one volt is applied across it.
4. The kilowatt-hour (kWh), also known as the Board of Trade unit, is the commercial unit of electrical energy (B.O.T).
5. 1kwh =1000×60×60 watt-second 1kwh=3.6×10⁶W sor Joules. Generally, one kWh is called one unit.
Electric power:
1. Electrical power is defined as the rate at which an electrical circuit transfers electrical energy per unit of time. Electrical energy can be either kinetic or potential energy in this context.
2. In most cases, potential energy is taken into account, which is the energy stored as a result of the relative positions of charged particles or electric fields. P denotes electrical power, which is measured in Watts.
3. SI Unit is Watt, joule per second and Formula, P=VI Where, V is the potential difference (volts), I is the electric current.
(ii) ) Energy consumed by 2 bulbs/day = (2 x 50 x 6) = 600W - h = 0.6kWh
Total energy consumption/day = 0.6 + 1 = 1.6kWh
Total energy consumption/month = (1.6 x 30) = 48kWh,
Total cost = 48 x 8 = Rs. 384

According to Ohm's law,
R=V/I
Here, R=10/1=10 Ω
Since the heater and conductor are in series, resistance of electric lamp = Total resistance- Resistance of the conductor
Hence resistance= 10−5=5 Ω
Now, when a resistance of 10 Ω is connected in parallel with this series combination, the total resistance is reduced to half of the original value and hence the total current is double the original value.
Therefore, there will be no change in the current flowing through 5Ω conductors and the potential difference across the lamp will also remain the same.

(i) \(R_{A C}\) and \(R_{E D}\) are in parallel, so
\(\frac{1}{R_p^{\prime}}=\frac{1}{R_{A C}}+\frac{1}{R_{E D}}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}\)
\(\Rightarrow \quad R_p^{\prime}=15 \Omega\)
Now, \(R_p\) and \(R_{B C}\) are in series, so
\(R_s  =R_p^{\prime}+R_{B C} =15+15=30 \Omega\)
Again, \(R_{A B}\) and \(R_s\) are in parallel, so
\(\frac{1}{R_p^n}  =\frac{1}{R_{A B}}+\frac{1}{R_S^{\prime}} =\frac{1}{15}+\frac{1}{30}=\frac{1}{10}\)
\(\therefore \quad R_p^n  =10 \Omega\)
So, current flowing through the circuit,
\(I=\frac{V}{R_p^{\prime \prime}}=\frac{3}{10}=0.3 \mathrm{~A}\)
(ii) Two 8Ω resistors are connected in parallel. Two such parallel combination must be connected in series to get effective resistance of 8 Ω.
Such combination is shown as below 

(i) The experimental set up comprise three resistors R₁,R₂ and R₃ of three different values such as 1 Ω, 2 Ω and 3 Ω which are connected in series.
Connect them with a battery of 6 V , an ammeter and plug key, as shown in figure

The key K is closed and the ammeter reading is recorded. Now, the position of ammeter is changed to anywhere in between the resistors again, the ammeter reading is recorded each time. It's observed that there was identical reading each time, which shows that same current flows through every part of the circuit containing three resistances in series connected to a battery.
(ii) (a) Equivalent resistance of given circuit is R, then
\(R  =(24 \| 24)+12=\frac{24 \times 24}{24+24}+12\) =12+12=24 Ω
Therefore Current through $12 Ω resistor,
\(I=\frac{V}{R}=\frac{6}{24}=0.25 \mathrm{~A}\)
(b) Difference in reading of A₁ and A₂ = \((0.25-0.25) \mathrm{A}=0 \mathrm{~A} \)

Equivalent resistance of the given network is
\(\overset { 1 }{ \underset { R }{ \_ \_ } } =\overset { { V } }{ \underset { { R }_{ 4 } }{ \_ \_ } } +\overset { 1 }{ \underset { { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 } }{ \_ \_ \_ \_ } } =\overset { 1 }{ \underset { 10 }{ \_ \_ } } +\overset { 1 }{ \underset { 10+10+10 }{ \_ \_ \_ \_ } } \)
\(\overset { 1 }{ \underset { 10 }{ \_ \_ } } +\overset { 1 }{ \underset { 30 }{ \_ } } =\overset { 3+1 }{ \underset { 30 }{ \_ \_ } } =\overset { 4 }{ \underset { 30 }{ \_ \_ } } \)
\(\therefore\) R = \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 3 }{ \underset { 7.5 }{ \_ \_ } } =7.5\Omega \)
Current drawn from the battery
I =\(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 3 }{ \underset { 7.5 }{ \_ \_ } } =\overset { 30 }{ \underset { 75 }{ \_ \_ } } =\overset { 2 }{ \underset { 5 }{ \_ \_ } } \)
I = 0.4 A

 (i) (a) In Parallel  \(I=\overset { V }{ \underset { R }{ \_ \_ } } \) or 2 x 5 = V = 10 V
For I0  \(\Omega \) resistance I =  \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 10 }{ \underset { 10 }{ \_ \_ } } \) =1A 
For 15 \(\Omega \) resistance I =  \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 10 }{ \underset { 10 }{ \_ \_ } } \) = 0.7A 
(b)  \(I=\overset { V }{ \underset { R }{ \_ \_ } } \)  
or, 2 =\(\overset { V }{ \underset { 5 }{ \_ \_ } } \)
or V=10 V
(c) Resistance in Parallel
\(\overset { 1 }{ \underset { R }{ \_ \_ } } =\overset { 1 }{ \underset { 10 }{ \_ \_ } } +\overset { 1 }{ \underset { 15 }{ \_ \_ } } =\overset { 1 }{ \underset { 30 }{ \_ \_ } } =\overset { 1 }{ \underset { 6 }{ \_ \_ } } \)
So, R= 6 \(\Omega \)
So, Total resistance across AC
= 5 + 6 = 11\(\Omega \)
 (ii)
 
 If we connect all the resistance in parallel we will get the least total resistance.

(a) Since two \(8\Omega \) resistors are in parallel, their effective resistance (\({ R }_{ P }\)) is given by,
\(\frac { 1 }{ { R }_{ P } } =\frac { 1 }{ 8 } +\frac { 1 }{ 8 } +\frac { 1 }{ 4 } \) or \({ R }_{ P }=4\Omega \)
(b) Total resistance in the circuit
\(R=4\Omega +{ R }_{ P }=4\Omega +4\Omega =84\Omega \)
Current through the electric circuit
\(I=\frac { V }{ R } =\frac { 8V }{ 8\Omega } =1A\)
(c) The potential difference \(4\Omega \) resistor V
 I \(=IR=1\times 4=4V\) 
(d) Power dissipated in \(​​​​4\Omega \) resistor P=\({ (I) }^{ 2 }\)
\(R={ (1) }^{ 2 }(4)=4W\)
(e) There is no difference in the readings of ammeters \({ A }_{ 1 }\) and  \({ A }_{ 2 }\) as same current flows through all elements in a series circuit.

Resistors in series: Join 3 resistors R₁, R₂ and R₃ of three different values in series as shown in the figure. Connect them with battery, an ammeter, a plug key. Use resistors of value 1 Ω, 2 Ω, 3 Ω etc, and a battery of 6 V for performing this activity.
Note the ammeter reading.
Change the position of ammeter to anywhere in between the resistors.

Observations: The value of current I in the ammeter is same. Let it be placed anywhere in the circuit.
Conclusion: In series combination of resistors, the value of current is the same in every part of the circuit.

(i)  (b) 4 Ω
(ii) (c) 1 A
(iii) (c) 4 V 
(iv) (c) 4 W
(v)  (d) No difference

(i)  (c) 24 Ω
(ii)  (a) 0.25 A
(iii)  (a) 5 V
(iv)  (d) 1 V
(v)   (b) 1.25 W

(i)  (b) IR²
(ii) (c) 484 Ω
(iii) (d) 25 W
(iv) (c) kilowatt-hour
(v)  (b) 0.18 A

(i) (d) 220 V
The potential different between live and neutral wire is 220 Volts
(ii) (c) Live wire 
Switches are connected in the live wire because when the switch is in the off position, no point of the connected electrical appliance will be at higher potential (220 V). 
(iii) (a) 15 A 
A fuse of rating 15 A is usually used for appliance electric iron, geysers and room heater etc.
(iv) (a) Earthing 
The earthing of any electrical appliance is done to protect the user from any accidental electrical shock due to leakage of current.
(v) (b) in parallel circuit if one device is damaged, then it does not affect other devices.

R₂ and R₃ are in series, thus for this combination,
\(R^{\prime}=R_2+R_3\)
Similarly, R₄ and R₅ are in series.
So, \(R^{\prime \prime}=R_4+R_5\)
K' and \(R^{\prime}\) are in parallel.
R₁ and R''' are in series. 
\(\therefore R_{\text {eq }}=R_1+\frac{\left(R_2+R_3\right)\left(R_4+R_5\right)}{R_2+R_3+R_4+R_5}\)
\(\therefore \quad R^{\prime \prime}=\frac{R^{\prime} R^{\prime \prime}}{R^{\prime}+R^{\prime}} \)
\(R^{\prime \prime \prime}=\frac{\left(R_2+R_3\right)\left(R_4+R_5\right)}{R_2+R_3+R_4+R_5}\)

(a) Ohm's law states that the current flowing through a metallic conductor is directly proportional to the potential difference applied across the ends of the metallic conductor, provided the physical conditions remain the same, i.e., temperature.
Equivalent resistance \(R_p\) of the parallel combination of three resistor of values \(R_1, R_2\) and \(R_3\) is given by
\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
(b) Across the length \(X Y\) of wire, three resistors lies between the wire.

Resistors 1 and 2 are in parallel combination,
\( \frac{1}{R}=\frac{1}{R}+\frac{1}{R}=\frac{2 R}{R^2}\)
\(\Rightarrow  R^{\prime}=\frac{R}{2}\) and \(R^{\prime}\) and resistor 3 are in series combination
\(R^{\prime \prime}=R+R=\frac{R}{2}+R=\frac{3 R}{2}\)