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**#Technical Studies – Study 26**

**#Technical Studies – Study 26**

**Introduction**

**Introduction**

Electric current always flows in a closed path. That closed path is known as electric circuit. A circuit in which direct current flows is called as dc circuit. In a dc circuit there are number of components connected, which are known as loads. Each load in a dc circuit has some resistance. So each load can be considered as a resistance. According to the connection of resistances in a dc circuit, it can be classified into two types – series and parallel. The combination of these two types can be considered as another one type of dc circuit i.e. series – parallel circuit.

*Series circuit:*

*Series circuit:*

In this type of circuit the resistances or loads are connected in series. That is one end of a resistance is connected to the end of another resistance and so on. In a series circuit current has only one path to flow.

Consider the below circuit. There are four resistances connected end to end across a voltage source V. current I is flowing through the single path and it is same for all the loads in the circuit. But voltage is different for different loads.

According to ohm’s law V = IR. But this circuit has four resistances and the voltage drops are different for different resistances.

So, Voltage drop across ${R}_{1}={V}_{1}=I{R}_{1}$

Voltage drop across ${R}_{2}={V}_{2}=I{R}_{2}$

Voltage drop across ${R}_{3}={V}_{3}=I{R}_{3}$

Voltage drop across ${R}_{4}={V}_{4}=I{R}_{4}$

So, the total voltage of the circuit $V={V}_{1}+{V}_{2}+{V}_{3}+{V}_{4}$

$=I{R}_{1}+I{R}_{2}+I{R}_{3}+I{R}_{4}$

$=I({R}_{1}+{R}_{2}+{R}_{3}+{R}_{4})$

$\frac{V}{I}={R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}$

As per ohm’s law the total resistance of the circuit ${R}_{s}=\frac{V}{I}$

Here ${R}_{s}$ is the total resistance of series circuit.

Therefore ${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}$

So in a series circuit the total resistance of the circuit is equal to the sum of individual resistances.

The conductance of a series circuit can be $G=\frac{1}{{R}_{s}}=\frac{1}{{R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}}$

The power rating of the circuit is

$P=VI$

$=({V}_{1}+{V}_{2}+{V}_{3}+{V}_{4})I$

$=(I{R}_{1}+I{R}_{2}+I{R}_{3}+I{R}_{4})I$

$={I}^{2}{R}_{1}+{I}^{2}{R}_{2}+{I}^{2}{R}_{3}+{I}^{2}{R}_{4}$

$={P}_{1}+{P}_{2}+{P}_{3}+{P}_{4}$

The total power dissipated in the circuit is the sum of the power dissipated by individual resistances.

*Parallel circuit:*

*Parallel circuit:*

In parallel circuit the resistances or the loads are connected in parallel configuration. That means the entire circuit has two common points where both the ends of all the resistances are connected.

Consider the below circuit. There are four resistances $({R}_{1},{R}_{2},{R}_{3},{R}_{4})$ connected in parallel across a battery of voltage V, and current I is flowing in four different paths. Voltage V is same for all the resistances.

According to ohm’s law V = IR. But this circuit has four resistances in parallel and current flowing through each resistance is different.

So, Current flowing through resistance ${R}_{1}={I}_{1}=\frac{V}{{R}_{1}}$

Current flowing through resistance ${R}_{2}={I}_{2}=\frac{V}{{R}_{2}}$

Current flowing through resistance ${R}_{3}={I}_{3}=\frac{V}{{R}_{3}}$

Current flowing through resistance ${R}_{4}={I}_{4}=\frac{V}{{R}_{4}}$

So the total current flowing in circuit $I={I}_{1}+{I}_{2}+{I}_{3}+{I}_{4}$

$=\frac{V}{{R}_{1}}+\frac{V}{{R}_{2}}+\frac{V}{{R}_{3}}+\frac{V}{{R}_{4}}$

$=V(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}})$

$\frac{I}{V}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}}$

We know that $R=\frac{V}{I}$

So, $\frac{I}{V}=\frac{1}{R}$

The total resistance of parallel circuit is denoted as ${R}_{p}$

Therefore $\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}}$

So in a parallel circuit the reciprocal of total resistance is equal to the sum of reciprocal of individual resistances.

Power dissipated in a parallel circuit

$P=VI$

$=V({I}_{1}+{I}_{2}+{I}_{3}+{I}_{4})$

$=V(\frac{V}{{R}_{1}}+\frac{V}{{R}_{2}}+\frac{V}{{R}_{3}}+\frac{V}{{R}_{4}})$

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

$={P}_{1}+{P}_{2}+{P}_{3}+{P}_{4}$

The total power dissipation of the circuit is the sum of the power dissipated by each resistance.

**Series-parallel circuit:**

**Series-parallel circuit:**

This type of circuit is the combination of both series and parallel circuits.

Consider the below circuit. In this circuit ${R}_{2}$$,{R}_{3}$ and ${R}_{4}$ are connected in parallel to each other and they all are series with ${R}_{1}$. To calculate the total resistance of the circuit, we need to calculate the total resistance of the parallel branch first and then calculate the total resistance in series.

Total resistance of the circuit $={R}_{1}+(\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}})$

*Problems:*

*Problems:*

*Question 1: *

*Question 1:*

There are three resistors connected in series across 25 V supply. if the value of three resistors are 50 Ω, 40 Ω, 70 Ω.

(i) Calculate the current of the circuit.

(ii) Calculate the voltage across each resistance.

(iii) Calculate the total power of the circuit.

Solution:-

Given – ${R}_{1}=50$Ω

${R}_{2}=40$Ω

${R}_{3}=70$Ω

V = 25 V

(i) Resistors are connected in series so the total resistance ${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}$

= 50 + 40 + 70 = 160 Ω

Therefore the circuit current $=\frac{V}{{R}_{s}}=\frac{25}{160}=0.156A$

(ii) Voltage drop across ${R}_{1}=I{R}_{1}=0.156\times 50=7.8V$

Voltage drop across ${R}_{2}=I{R}_{2}=0.156\times 40=6.24V$

Voltage drop across ${R}_{3}=I{R}_{3}=0.156\times 70=10.92V$

(iii) Total power dissipated $P={P}_{1}+{P}_{2}+{P}_{3}$

$={I}^{2}{R}_{1}+{I}^{2}{R}_{2}+{I}^{2}{R}_{3}$

$={(0.156)}^{2}\times 50+{(0.156)}^{2}\times 40+{(0.156)}^{2}\times 70$

= 1.2+0.96+1.68

= 3.84 watt

*Question 2:*

*Question 2:*

There are three resistors 6 Ω, 8 Ω and 14 Ω connected in parallel.If the circuit current is 10 A, find the current through each resistor.

Solution:-

Given resistor values – 6 Ω, 8 Ω, 14 Ω

$\frac{1}{{R}_{p}}=\frac{1}{6}+\frac{1}{8}+\frac{1}{14}=\frac{28+21+12}{168}=\frac{61}{168}$

Therefore ${R}_{p}=\frac{168}{61}=2.754$

Voltage across the circuit = $I{R}_{p}$ $=2.754\times 10=27.54V$

Current flowing through 6 Ω resistor $=\frac{27.54}{6}=4.59A$

Current flowing through 8 Ω resistor $=\frac{27.54}{8}=3.44A$

Current flowing through 15 Ω resistor $=\frac{27.54}{14}=1.96A$

*Question 3: *

*Question 3:*

See the below circuit:-

(i) Calculate the total resistance of the circuit.

(ii) Calculate the conductance of the circuit.

Solution:-

In the given circuit ${R}_{1}=2$ Ω

${R}_{2}=5$ Ω

${R}_{3}=3$ Ω

${R}_{2}$ and ${R}_{3}$ is connected in parallel and they are in series with ${R}_{1}$

So the total resistance of the parallel branch = $\frac{1}{{R}_{p}}=\frac{1}{5}+\frac{1}{3}=\frac{3+5}{15}=\frac{8}{15}$

Therefore ${R}_{p}=\frac{15}{8}=1.875$Ω

Now the series resistance = 2+ 1.857 = 3.875 Ω

Conductance of the circuit G = $\frac{1}{R}=\frac{1}{3.875}=0.258S$

Please write in the comment box below if you have any questions.