As we know that in the series circuit the current remain the same and in the parallel circuit the voltage remains the same.

WHEN THE CELL ARE CONNECTED IN SERIES

Now consider the series circuit in which all the cell are connected in series. If there is “n” number of the cell than the total EMF is nE.

Now all the battery has total internal resistance therefore if n cell is connected in series that the total internal resistance will be “Rn”

Now total current in the circuit is I = nE/nR = E/R

WHEN THE CELL ARE CONNECTED IN PARALLEL

When n number of cell are connected in parallel then total EMF is equal to E (since in parallel circuit voltage remain the same).

Internal resistance in parallel = R’ = R/n

Total current in parallel circuit = I = nE/R

Now consider 3 cell each of 2V having the internal resistance of 0.1 ohm

Current in series = I = E/R = 2/0.1 = 20A

Current in parallel = I = nE/R = 60A

Hence when the individual cells are connected in parallel, the voltage is the same as that across one cell. However, the current rating of the combination equals the sum of the individual current-rating values. Only cells that have the same voltage should be connected in parallel. Whereas when the cells are connected in a series-aiding manner. The current rating of the series-aiding cells is the same as that for the cell with the lowest current rating.

Let the equivalent resistance when they are connected in parallel be R_P

1/R_P = 1/R + 1/R ———– N times

1/R_P = N/R

R_P = R/N

Since the equivalent resistance, R_P is given as X

∴ X = R/N

or

R = NX

Now the equivalent resistance when they are connected in series be R_s

R_s = R + R———- N times

R_s = NR

R_s = N²X-————————–(since R = NX)

Hence the total resistance when these N resistances are connected in series is R_s = N²X

Electrical Power

The rate at which the electric work is done in an electric circuit is called an electric power. The SI unit of power is joule per second or watt.

∴ 1 watt is defined as 1 joule/second.  i.e
1 watt – sec = 1 joule

So 1kW is 1000W, and an hour is 60 seconds x 60 minutes.

So 1kWh = 3600kWs = 3,600,000Ws = 3,600,000J

0.5 kWh = 0.5 × 3,600,000Ws = 1800000 joules

Given

Length =10 m

Radius = Diameter/2 = 0.8/2 =0.4m

Area of circle = πR² = π × 0.4²

Resistance = 2Ω

The formula of resistivity is given as

∴ 

$\begin{array}{l}\rho = \dfrac{{\pi \times {{0.4}^2} \times 2}}{{10}}\\\\\rho = 0.1\Omega – m\end{array}$

Conductivity = 1/Resistivity

G = 1/0.1 = 10 Mhos/m

Erg is the unit of energy or work in the centimeter-gram-second system. It is equivalent to 1.0 × 10⁻⁷ Joules or 2.4 × 10⁻⁸ calories.

Let us mark the capacitance as shown in the figure

The capacitance C₂, C₃, C₄ are in parallel with capacitance C₆, C₇

1/C = (20 + 20 + 10) || (30 + 20)

C = (1/20 + 1/20 + 1/10) || (1/30 + 1/20)

C = 5 + 12 = 17 μF

Now all the three capacitors are connected in series as shown in the figure.

C = 17 + 17 + 17

1/C = 1/17 + 1/17 + 1/17

C = 5.67μF

Let us mark the capacitance as shown in the figure

The inductance L₂, L₃, L₄ are in series therefore equivalent inductance

L = 1 + 1 + 1 = 3H

Similarly the inductance L₆, L₅ are connected in series 

L = 1 + 1 = 2 H

Now the inductance 3H and 2H are connected in parallel with each other therefore

L = 3 || 2 = (3 × 2) ⁄ (3 + 2)

L = 1.2 H

Now the all the three inductance are connected in series hence

L = 1 + 1.2 + 1

L = 3.2 H

Capacitors with Dielectrics

A dielectric is an insulating material such as rubber, glass, Teflon or waxed paper. Let us assume the capacitor is charged but not connected to anything.

Now, if the air space between the plates is filled with the dielectric medium such as Teflon the magnitude of the electric field falls down. Weakening in E means decrement in V, since E = V/d. Q remains constant, because there is no wire to carry the charge so it does not have a path to flow. Now the decrement in V will mean an increment in the value of C (using Q = CV).

Note:-  If the capacitor remains connected to the battery, the voltage across the capacitor necessarily remains the same. If you disconnect the capacitor from the battery before making any modifications to the capacitor, the capacitor is an isolated system and its charge remains the same because the capacitor discharges when a conducting path is provided across the plates without any applied voltage connected.

Resistor values can be known by the color band.

0 black

1 brown

2 red

3 orange

4 yellow

5 green

6 blue

7 violet

8 grey

9 white

First color = digit
Second color=digit
Third color= power of ten
Fourth color =tolerance

Now in the given question, the color of the resistor is given as brown-black-brown-black

1 = Brown

0 = Black

10¹ = Brown

± 0% = black

We get Resistance value from the color code (Brown,Black,Brown,Black) is 10 x (10¹) = 100 ohm

Here we are not considering the last color (tolerance-black) Because in the question there is no specification of minimum or maximum current.

Electrolytic Capacitors are generally used in DC power supply circuits due to their large capacitance and small size to help reduce the ripple voltage or for coupling and decoupling applications. Electrolytic’s generally come in two basic forms; Aluminium Electrolytic Capacitors and Tantalum Electrolytic Capacitors. The tantalum and electrolytic capacitors have ratings of 6, 10. 12. 15. 20. 25, 35, 50. 75, I00V, and higher.

Paper capacitors: Paper capacitors use paper as the dielectric. They are made of flat strips of metal foil plates separated by a dielectric, which is usually waxed paper. Paper capacitors have values in the picofarad and low microfarad, ranges. The voltage ratings are usually less than 6 V. Paper capacitors arc usually sealed with wax to prevent moisture problems.

The voltage rating of capacitors is very important. A typical set of values marked on a capacity might be “10 μF, 50 DCWV.” Such a capacitor h a capacitance of 10 μF and a “dc working voltage” of 50V. This means that a voltage in excess of 50 could damage the plates of the capacitors.

Ceramic capacitors: Ceramic capacitors make use of ceramic as the dielectric material. Basically, ceramic materials are fanned from titanium dioxide. Some of the dielectric materials used are barium titanate, magnesium titanic, and zirconium titanate.

Note:-

Wire Wound Resistors:- The wire wound resistors have a resistive wire wound on a central ceramic core. It is one of the oldest technologies, wire wounds provide the best-known characteristics of high-temperature stability and power handling ability. Nichrome, Manganin, and Evanohm are the three most widely used wires for wire wound resistors. Wirewound resistors have a significant amount of inductance due to the construction technique.