Given, Resistance, R = 8 \(\Omega\) , Current, J = 15 A
Time, t = 2 h = 7200 s
\(\therefore\) Heat developed. H = J²Rr
=15 \(\times\)15 \(\times\) 8 \(\times\) 7200 J
\(\therefore\) Rate of heat developed.
\(P=\frac{H}{t}=\frac{15\times 15\times 8 \times 7200}{7200}\)
= 1800 W or 1800J/s
Thus, the rate at which heat is developed in the heater is 1800 joule per second.

(i) If two 6 \(\Omega\) resistors are connected in parallel, then the equivalent resistance is \(\left ( \frac{6}{2} \right )=3 \Omega \)
This combination is connected in series with a 6 \(\Omega\) resistor to get overall equivalent resistance of (6+3) =9 \(\Omega\).
(ii) Equivalent resistance of two 6 \(\Omega\) resistances connected in series, R'=6+6= 12 \(\Omega\). Now, 12 \(\Omega\) and 6 \(\Omega\) resistors are connected in parallel.
Equivalent resistance, \(R_{eq}=\frac{12\times 6}{12+6}=\frac{72}{18}=4 \quad \Omega \)

As the wire is bent in the form of a closed circle, it will be have like two resistors of 5\(\Omega \) each, connected in parallel. Therefore, the equivalent resistance can be calculated as:
\(\frac { 1 }{ R_{ eq } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ R_{ eq } } =\frac { 1 }{ 5 } +\frac { 1 }{ 5 } \)=\(\frac { 1 }{ R_{ eq } } =\frac { 2 }{ 5 } \)
\(\frac { 1 }{ R_{ eq } } =\frac { 5 }{ 2 } =2.5\Omega \)
The effective resistance between the two points at the ends of an diameter of the circle is 2.5 \(\Omega \).

(a) All the appliances work at the same voltage as that of the electric supply.
(b) If one of the other appliances is out of order, all the other appliances keep on working as the circuit is not broken.

Ohm (1Ω)

A measuring instrument used to measure potential difference between two points in an electric circuit.

A measuring instrument used to measure electric current.

Electrical power determines the rate at which the energy is delivered by a current

It means that 1 joule of work will be done in order to move 1 coulomb of charge from point A to B.

(ii) When the key is closed, total voltage applied across the circuit will be = 2 ✕ 5V = 10 V
     Total resistance (R) applied across the circuit in Ohms = 5+10+15= \(30\Omega \) 
     By Ohm's Law,
     Current (I) = \(\frac { V }{ R } \)
       \(I=\frac { 10 }{ 30 } =0.33A\)

(a) Ohm's Law states that at constant temperature, current flowing through a conductor is directly proportional to the       potential difference across its ends.

(a) Short circuiting is the term used for defining the situation in which the neutral and live wires of an electric circuit come in direct contact. When too many electrical appliances are connected to a single socket, the appliances drawn much more current or power that the permissible limit from the main supply. This situation is termed as overloading.
(b) Resistivity of an alloy is much higher than its constituent metals Moreover, alloys do not oxidise as easily as the constituent metals at high temperature. Because of these reasons, coils of electric testers are made of an alloy rather than a pure meta.

(a)
  
(b) Electric current flowing through a bulb = \(\frac { Power\quad of\quad the\quad bulb }{ Voltage\quad across\quad the\quad bulb } \)
\(i.e.,\quad I=\frac { P }{ V } \)
Therefore, \({ I }_{ 60 }=\frac { 60W }{ 220V } =\frac { 3 }{ 11 } A\quad and\quad { I }_{ 40 }=\frac { 40W }{ 220V } =\frac { 2 }{ 11 } A\)
Therefore, total current flowing through the circuit  \(={ I }_{ 60 }+{ I }_{ 40 }=\frac { 3 }{ 11 } +\frac { 2 }{ 11 } =\frac { 5 }{ 11 } A=0.45A\)
(c) Total energy consumed = Power \(\times \) Time
      That is, E = P \(\times \) T
                      = (40W+60W)\(\times \)1h
                    = 100 Wh
                    = 0.1 kWh

(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.

A number of resistors are said to be connected in series, if these are joined end to end and the same, current flows through each one of them when a potential difference is applied across the combination. If we connect ammeter first between the point A and R₁, then between R1 and R₂:R₂ and R₃ and lastly between R₃ and the point B; it is same as current.