# How to Calculate Resistors in Series and Parallel: 30+ Solved Examples, Circuit diagram

Table of Contents

## Introduction:

**How to Calculate Resistors in Series and Parallel: 30+ Solved Examples, Circuit diagram-** in this article, you will learn how to calculate the resistance of resistors connected in series and parallel. As an electronics student or as an engineering student, you should be able to solve simple and complex circuits. We will start with simple circuits and then in the end we will also solve some complex circuits as well. We will also solve circuits consisting series and parallel connections.

## Calculations about Resistors in Series Circuits

**Example 1;** Three resistors of 2-ohm, 3-ohm, and 6-ohms are connected in series across a 4-volt supply. Find;

(a). Total resistance (b). Current (c). Voltage drops across each resistor

**Solution;**

Let R_{1 }= 2 Ω, R_{2} = 3 Ω and R_{3 }= 6 Ω

(a). If RT is the total resistance of the circuit, then

R_{T} = R_{1} + R_{2} + R_{3} = 2 +3 + 6 = 11 Ω

(b). If I is the current flowing through the circuit, then by applying ohm’s law;

I = V_{T }/ R_{T} = 4/ 1_{1 }= 0.363 Amp.

©. Voltage drops across R_{1}, V_{1} = IR_{1 }= 0.363 x 2 = 0.727 V

Voltage drops across R_{2}, V_{2} = IR_{2} = 0.363 x 3 = 1.089 V

Voltage drops across R_{3}, V_{3} = IR_{3} = 0.363 x 6 = 2.178 V

**Example 2;** What is the total resistance of four resistors connected in series if their individual resistance values are 1M Ω, 1.5M Ω, 150K Ω, and 50000 Ω.

**Solution;**

R_{T} = R_{1} + R_{2} + R_{3 }+ R_{4}

R_{T} = 1M Ω + 1.5M Ω + 0.15M Ω +0.05M Ω = 2.7M Ω

∴ 1M Ω = 106 Ω

Figure 1.34

**Example 3;** A series circuit consisting of three resistors, 2, 8, and 20 Ω, connected to a battery has a current of 2A. what voltage exists across each resistor and also calculate the total voltage of the battery.

**Solution;**

V_{1} = I R_{1} = 2 x 2 = 4 V

V_{2} = I R_{2} = 2 x 8 = 16 V

V_{3} = I R_{3} = 2 x 20 = 40 V

Total voltage V= V_{1 }+ V_{2} + V_{3 }= 4 + 16 + 40 = 60 V

Figure 1.35

**Example 4:** What value resistor connected in series with a 1.8- KΩ resistor, will result in a total resistance of 2KΩ.

**Solution;**

We know that;

R_{T} = R_{1 }+ _{R2}

R_{2} = R_{T }– R_{1} = 2 KΩ – 1.8 KΩ = 0.2 KΩ or 200 Ω

Figure 1.36

**Example 5;** Three resistors are connected in series. The values of the two resistors are 10 KΩ and 20 KΩ. The supply voltages are 200 volts. If the current flowing through the third resistor is 5mA, find the value of this resistor.

**Solution;**

V_{T} = 200 V, R_{1 }= 10 KΩ, R_{2} = 20 KΩ

R_{3} =? ; I = 5mA

In series circuit, the current flows in all resistance is same i.e., 5mA, so

V_{1 }= 5mA x 10 KΩ = 50 V

V_{2} = 5mA x 20 KΩ = 100 V

Sum of V_{1} and V_{2} = 50 + 100 = 150 V

But 150 + V_{3 }must be equal to 200 V, so

V3 = 200 – 150 = 50 V

Now 50 V= I. R_{3 }or R_{3 }= 50 V / I = 50 V / 5mA = 50 / 0.005 = 10000 Ω or 10 KΩ

**Example 6;** Three resistances of 5 KΩ, 10 KΩ, and 15 KΩ are connected in series across a 24-voltage source. Using voltage division rule, determine voltage drops across each resistance.

**Solution;**

V = 24 V

Let R_{1} = 5 KΩ, R_{2 }=10 KΩ, and R_{3 }= 15 KΩ

Now total resistance of the series circuit is

R_{T} = R_{1} + R_{2} + R_{3} = 5 + 10 + 15 = 30 KΩ

According to voltage division rule, various voltage drops are;

V_{i} = V x R_{1} / R_{T} = 24 x 5 / 30 = 4 V

V_{2 }= V x R_{2} / R_{T }=24 x 10 / 30 = 8 V

V_{3} = V x R_{3 }/ R_{T }= 24 x 15 / 30 = 12 V

## Calculations About Resistors in Parallel Circuits

**Example 7;** Three resistors of 4 ohms, 6 ohms, and 8 ohms respectively are connected in parallel. Find the equivalent resistance.

**Solution;**

1 / R_{e} + 1 / R_{1} + 1 / R_{2 }+ 1 / R_{3}

= 1 4 + 1 /6 + 1 / 8 = 6 + 4 + 3 / 24 = 13 / 24

Re = 24 / 13 = 1.846 ohms.

**Example 8;** Two resistors 3.1 ohms and 7.2 ohms respectively are connected in parallel. Find the equivalent resistance.

**Solution;**

Equivalent resistance = Product / Sum

= 3.1 x 7.2 / 3.1 + 7.2 = 22.32 / 10.30 = 2.16 ohms

**Example 9;** A battery with a 12 V emf and negligible internal resistance form a parallel circuit with a network of three resistors, 2, 4, and 6 ohms. Calculate

(a). Total resistance (b). Current in each resistor (c). Total current

**Solution;**

Given V = 12 V, R_{1 }= 2 Ω, R_{2} = 4 Ω, and R_{3} = 6 Ω

Figure 1.37

Total resistance

1 / R_{T} = 1 /R_{1 }+ 1/ R_{2 }+ 1/ R_{3 }= ½ + ¼ + 1/6 = 6 + 3 + 2 / 12 = 11 / 12

R_{T} = 11 / 12 = 1.09 Ω

(b). The 12 V battery is connected to the three resistors R1, R2, and R3. The current in each resistor is related to that voltage by ohm’s law;

Current through resistor R_{1}, I_{1} = V / R_{1 }= 12 / 2 = 6A

Current through resistance R_{2}, I_{2} = V / R_{2} = 12 / 4 = 3A

Current through resistance R_{3}, I_{3 }= V / R_{3} = 12 / 6 = 2 A

(c).

(i) Total current I_{T}= I_{1} + I_{2 }+ I_{3 }= 6 + 3 + 2 = 11A

(ii). I_{T} = V / R_{T} = 12 / 1.09 = 11 A

**Example 10;** Calculate the total equivalent resistance of the parallel network shown below;

Figure 1.38

**Solution;**

1 / R_{T} = 1 / R_{1 }+ 1 / R_{2 }+ 1 / R3 + 1 / R4

= 1 /12 + 1 / 15 + 1/ 18 + + 1/33

= 0.0833 + 0.0667 + 0.0556 + 0.0303

= 0.2359

R = Reciprocal of 0.2359 = 4.24 Ω

**Example 11;** A generator is applied to a parallel network consisting of four resistors 10, 20, 25, and 50 Ω. The current in the 25 Ω resistor is 4A. what are the other circuit currents.

Figure 1.39

**Solution:**

Since this is a parallel network, there is only one circuit voltage and so the voltage across R3 must be the same as the generator voltage.

V = I_{3} R_{3} = 4 x 25 = 100 V

This emf of 100 V will also be the voltage across the other three resistors R_{1}, R_{2}, and R_{4}.

I_{1} = V / R^{1} = 100 /10 = 10 A

I_{2 }= V /R_{2 }= 100 / 20 = 5 A

I_{4} = V /R_{4} = 100 / 50 = 2A

I_{T} = I_{1 }+ I_{2} + I_{3} + I_{4 }= 10 + 5+ 4+ 2 = 21A

**Example 12;** A parallel network has elements with resistances of 30, 40, 60, and 120 Ω. What is the network’s total equivalent

(a). Resistance (b). Conductance

Figure 1.40

**Solution;**

(a). 1 /R_{T }= 1 / R_{1} + 1 / R_{2 }+ 1/ R_{3} + 1 / R_{4 }= 1 /30 Ω + 1 /40 Ω + 1/ 60 Ω+ 1/ 120 Ω

=4 + 3+ 2+1 = 1/ 12 Ω

And R_{T} = 12 Ω

G_{T} = G_{1} + G_{2 }+ G_{3} + G_{4}

=1/30 S + 1/40 S + 1/60 S + 1/120 S

∴ S means Siemens

= 4 + 3+ 2+ 1 /120 = 1/12 S

Note; Siemens is the unit of conductance.

**Example 13;** what is the common voltage across the circuit in figure shown below;

Fig. 1.41

**Solution;**

Method 1; Applying ohm’s law to the entire circuit

V = I_{T} R_{T} = 5A x 8 Ω = 40V

Method 2; Applying ohm’s law to any one element;

V = I_{1}R_{1} =4A x 10 Ω = 40V

Or V = I_{2}R_{2} = 1A x 40 Ω = 40V

**Example 14;** What is the current I_{1} in the circuit shown below;

Fig. 1.42

**Solution;**

Method 1; ohm’s law (conductance form)

I_{1} = VG_{1 }= 40V x 0.1S = 4A

Method 2; Ohm’s law (resistance form)

I1 = V / R_{1} = 40V / 10 Ω = 4 A

Method 3; Resistance ration i.e., I_{1 }/ I_{2} = R_{2} / R_{1}

I_{1 }= I_{2} x R_{2 }/ R1

I_{1} = 1 x 40 / 10

=4A

**Example 15;** Two resistances of 4 ohm and 8 ohms are connected in parallel across a 6-volt source. Using current division rule, determine the amount of current flowing through each resistance.

**Solution;**

Given, V = 6 V, let R_{1} = 4 ohm and R_{2} = 8 ohm

Equivalent resistance is given by

R_{e }= R_{1}R_{2} / R_{1 }+R_{2} = 4 x 8/ 4 +8 = 32 / 12 = 2.66 Ω

Total current, I = V/ R_{e} = 6/ 2.66= 2.25A

Current passing through R_{1}

I_{1} = I [R_{2} /R_{1} + R_{2}] = 2.25 [8/ 4 + 8] = 1.5A

Similarly, current passing through R2

I_{2} = I [R_{1 }/ R_{1 }+ R_{2}] = 2.25 [4 / 4+8] = 0.75A

**Example 16;** Three resistors connected in parallel have an equivalent resistance 1.2 ohms. The value of two of the resistors are 6 ohms and 12 ohms. Find the value of third

**Solution;**

It is a parallel combination, so their equivalent resistance is;

1/Re = 1 / R_{1 }+ 1/ R_{2 }+ 1 /R_{3} + …

Let R3 be the unknown resistance, then

1 / 1._{2} = 1/6 + 1/12 + 1/R_{3} or 1/R_{3} = 1/ 1.2 – 1/6 -1/12

= 10 -2 – 1/ 12 = 7/12

R_{3} = 12 / 7 = 1.71 Ω

**Example 17;** Three resistors of 4, 12 and 6 ohms are joined in parallel. If the total current taken is 12 amperes, find the current through each.

Fig. 1.43

**Solution;**

The equivalent resistance of the parallel circuit is;

1/R = 1/4 + 1/12+ 1/6 = 3 + 1 + 2/ 12 = 6/ 12

R = 12/ 6 = 2 ohm

V = 1R = 12 x 2 = 24 V

Applying ohm’s law to each branch we get;

Current in 4 ohms resistor = V/R = 24/4 = 6A

Current in 12 ohms resistor = 24/ 12 = 2A

Current in 6-ohm resistor = 24/6 = 4A

**Example 18;** Three resistors of 2, 3 ohms and an unknown resistor x are joined in parallel. The current flowing through x is 5 amperes and the equivalent resistance of combination is 1 ohm. Find (a) value of the unknown resistor x (b) current through the other resistors (c) total current from the supply.

**Solution;**

In parallel combination, the equivalent resistance is given by;

(a}. 1/Re = 1/R1 + 1/ R2 + 1/ R3

1/1 = 1/2 + 1/3 + 1 /x or 1/x = 1/1 -1/2 – 1/3 = 6 – 3- 2 / 6 = 1/6 and x = 6 Ω

The current is flowing 5 amp. So, the voltage fed to the circuit is

V = 5 x 6 = 30 V

(b) Now current is 2 Ω = 30/2= 15A

Current in 3 ohms = 30/ 3 = 10 A

(c). In parallel combination, the total current is;

I = I_{1 }+ I_{2 }+ I_{3} = 15 + 10 + 5 = 30 A

**Example 19;** In the parallel arrangement of resistors shown the parallel current flowing in the 8 ohms resistor is 2.5 amperes. Find;

(a). Current in the other resistors (b). The resistor x (c). The equivalent resistance.

Fig. 1.44

**Solution;**

(a). The current in 8-ohm resistor is 2.5 A, so the voltage is.

=IR = 2.5 x 8 = 20 V

So current in 40- Ω resistor = 20 / 40 = 0.5 A

Current in 25- Ω resistor = 20/ 25 = 0.8 A

In parallel circuit, the total current is ;

I = I1 + I2 + I3 + I4

4 = 2.5 + _{2} + 0.5 + 0.8 and I2 = 0.2 A

(b). Now the resistance if taking 0.2 A at 20 V is V/ I = 20/ 0.2 = 100 Ω

(c). The equivalent resistance of the circuit, when taking 4A at 20V.

Re = V / IT = 20 / 4 = 5 Ω

## Calculations About Resistors in Series-Parallel Circuits

**Example 20;** For the arrangement shown, determine the voltage across the parallel branch and current in the main circuit.

Fig. 1.45

**Solution;**

The parallel combination may be replaced by a single resistor where;

1/R = 1/ 1 + 1/2 + 1/5 = 10 + 5 + 2 / 10 = 17 / 10

So, current in the circuit, I = V/ R_{T }= 10 / 5.588 = 1.79 A

Voltage across parallel branch = 1.79 x 0.588 = 1.05 V

**Example 21;** Find the resistance of the circuit shown.

Fig. 1.46

**Solution**

It is a compound circuit. Central branch has a parallel combination of 4-ohm resistances. So, the equivalent resistance of central parallel branch is;

Req = Product / Sum = 4 x 4 / 4 + 4= 16/ 8 = 2 Ω

So, the total resistance of the central branch is;

R = 3 + 2 = 5 Ω

Now we can say that there is a circuit having 2- Ω, 5 Ω, and 10 Ω resistances connected in parallel.

So, Req = 1/2 + 1/5 + 1/ 10 = 5 + 2+ 1 / 10 = 8/ 10

And Req = 10/ 8= 1.25 Ω

**Example 22;** Find the total resistance of the circuit shown.

Fig. 1.47

**Solution;**

Gave name to the circuit ABCD as shown. Let us calculate branch wise;

Branch AB.

Equivalent resistance of branch AB

R/ R_{AB} = 1/2 + 1/5 + 1/ 10 = 5 + 2 + 1/ 10 = 8/10

RAB = 10 / 8 = 1.25 Ω

Branch BC.

1/ R_{BC} = 1/6 + 1/4 = 2 + 3/ 12 = 5 / 12

R_{BC} = 12/5 = 2.4 Ω

Here, branch ABCD is a series circuit, so their total branch resistance;

R_{AB} + R_{B }+ R_{CD} = 1.25 + 2.4 + 1.35 = 5.00 Ω

Now we are having two resistances of 5 Ω connected in parallel. Their equivalent resistance is;

5 x 5/ 5 +5 = 2 /10 = 2.5 Ω

**Example 23;** Find the resistance of the circuit shown;

**Solution;**

Name the circuit and calculate branch wise. In branch ABCD, first solve parallel combination CD. In this section three resistance of 12 ohm are connected in parallel.

So, 1 / R_{CD} = 1/ 12 + 1/ 12+ 1/ 12 = 1 + 1 + 1 / 12 = 3 / 12

RCD = 12 / 3 = 4 Ω

BC Combination

1/ R_{BC} = 1/ 2 + 1 / 4 = 2 + 1 / 4 = 3/4

1/ R_{BC} = 1/2 + 1/4 = 2 + 1 / 4 = 3/4

R_{BC} = 4/3 = 1.33 Ω

Now both BC and CD are in series, their total resistance is;

R_{BD} = 4 + 4/3 = 16/3 Ω

And now R_{BD} and 3 Ω are in parallel, so;

1 / R_{BD} = 1/3 + 3/16 = 16 + 9 / 48 = 25 / 48

R_{BD} = 48/ 25 = 1.92 Ω

R_{AD} = R_{AB} + B_{D} = 2.08 + 1.92 = 4.00 Ω

Here are two branches, one of 16 Ω and other of 4 Ω is in parallel. So, their equivalent circuit;

1/ R_{eq} = 1/4’+ 1/16 = 4 + 1 / 16 = 5/16

R_{eq} = 16/ 5 = 3.2 Ω

**Example 24;** For the network shown in figure, the potential difference between A and C is 10 V. Calculate the current in each resistor.

Fig. 1.49

**Solution;**

R_{AB} = 6.0 x 2.0 / 6.0 + 2.0 = 1.5 Ω

R_{BC} = 7.0 x 3.0 / 7.0 + 3.0 = 2.1 Ω

R_{AC} = R_{AB} + R_{BC }= 1.5 + 2.1 = 3.6 Ω

I_{AC} = V_{AC }/ R_{AC} = 10 / 3.6 = 2.8 A

I_{6} = 2.0 / 6.0 + 2.0 x 2.8 = 0.7 A

I_{2 }= IAC – I6 = 2.8 – 0.7 = 2.1 A

I_{7} = 3.0 / 7.0 + 3.0 x 2.8 = 0.8 A

I_{3} = I_{AC} – I_{7} = 2.8 – 0.8 = 2.0 A

**Example 25;** Find the resistance of the circuit shown. If the total current taken by this circuit is 5A, what is the supply voltage?

Fig. 1.50

**Solution;**

Let R_{AB} = R, then;

1/R = 1/ 4 + 1/10 + 1/ 20 = 5 + 2+ 1/20 = 8/ 20

R = 20 /8 = 2.5Ω

R_{BC} =3/ 3x 2/ 3+ 2 = 6/5 = 1.2 Ω

R_{AD} = R_{AB} + R_{BC} + R_{CD} = 2.5 + 1.2 + 2.3 = 6 Ω

Equivalent resistance of the circuit = 6 x 6 / 6 + 6 = 36/ 12 = 3 Ω

Supply voltage V = IR = 5 x 3 = 15V

**Example 26;** In the series – parallel circuit shown, find (a) voltage control across 4-ohm resistor (b) the supply voltage

**Solution;**

The given circuit can be understood as under;

Fig 1.51 (a) and (b).

Now the current in 4 Ω is the sum of the currents flowing through 8 Ω and 10 Ω resistances. We know the voltage across 8 Ω and 10 Ω resistors is given as 50 Ω.

So, current in 8 Ω resistor, I = V/R = 50 / 8 = 6.25 A

Current in 10 Ω resistor =50/ 10 = 5 A

Sum = 6.25 + 5 = 11.25 V

This is the current flowing through the 4 Ω resistance.

(a). So, voltage drop across 4 Ω resistor = 11.25 x 4 = 45 V

The equivalent resistance of 8 Ω, 12 Ω, and 24 Ω resistors which are connected in parallel;

1/ R_{CD} = 1/8 + 1/12 + 1/24 = 6 + 4 + 2/ 48 = 12 / 48

R_{CD} = 48 / 12 = 4 Ω

Voltage drops across R_{CD} = 11.25 x 4 = 45 V

**Example 27;** Find the current supplied by the battery. Figure is shown below;

Figure 1.52

**Solution;**

Since R_{4} and R_{5 }are in parallel, we have;

1/ R_{6 }= 1/ R_{4} + 1/ R_{5} = 1/3 + 1/6 = 1 /2 and R_{6} = 2 Ω

Since R_{6} is in series with R_{3}, we have;

R_{7 }= R_{3 }+ R_{6} = 4 + 2= 6 Ω

Since R_{7} is parallel with R_{2}, we have;

1/8 = 1/ R_{7} + 1/ R_{2 }= 1/6 + 1/6 = 1/3 and R_{8} = 3 Ω

Since R_{8 }is in series with R_{1}, we have;

R_{9 }= R_{8 }+ R_{1} = 3 + 9 = 12 Ω

Current supplied by the battery is I = V /R_{9} = 6 / 12 = 0.5A

**Example 28;** What is the current in R_{3 }in figure shown;

Figure 1.53

**Solution;**

1/R_{A} = 1/R1 + 1/ R_{2}+ 1/ R_{3} = 1/12 + 1/6 + ¼ = 1/2 Ω and R_{A} = 2 Ω

1/ R_{B} =1/R_{4} + 1/ R_{5 }+ 1/ R_{6} + 1/R_{7} =1/96 + 1/ 48 + 1/16 + 1/32 = 1/8 Ω

R_{B} = 8 Ω

V_{A} = V [R_{A} / R_{A} + R_{B}] = 10 [2 / 2 + 8] = 2V

I_{3} = V_{A} / R_{3} = 2 Ω/ 4 Ω = 0.5A

**Example 29;** What is the voltage at point A to the ground in the circuit in figure given below;

Figure 1.54

**Solution;**

Rx is the parallel combination R_{1} and R_{2 }and R_{2} and Ry is the parallel combination R_{3}, R_{4}, and R_{5}.

R_{x }= R_{1} R_{2 }/ R_{1} + R_{2} = 12 KΩ x 22 KΩ/ 12 KΩ + 22 KΩ = 7.765 KΩ

1/R_{y} = 1/ R_{3} + 1/ R_{4} + 1/R_{5} = 1/ 18 KΩ + 1/ 33 KΩ + 1/27 KΩ = R_{7} = 8.13 KΩ

Applying the voltage – divider principle,

V_{y} = V_{T} [R_{y} / R_{x }+ R_{y}] = 15 [8.137 / 7.765 + 8.137] = 7.675 V

The potential at point A (with respect to ground) = +7.676 V

**Example 30;** What is the current IA in the circuit in figure shown

Figure 1.55

**Solution;**

R_{A }= R_{1 }+ R_{2} + R_{3} = 10 + 20 + 70 = 100 Ω

R_{B} = R_{4} + R_{5} + R_{6} = 50 + 150 + 300 = 500 Ω

Applying the current divider principle,

I_{A }= I_{T} [R_{B} / R_{A} + R_{B}] = 1 [500 / 100 + 500] = 0.833 A

**Example 31;** Determine I and the voltage across the 7 Ω resistor for the network of figure below;

Figure 1.56

**Solution;**

Network redrawn in fig. 1.37

R_{T} = (2) (4 Ω) + 7 Ω = 15 Ω

I = E / R_{T} = 37.5 Ω/ 15 Ω = 2.5 A Ans.

V _{7Ω} = I_{R} = (2.5 A) (7 Ω) = 17.5 V

Figure 1.57

**Example 32;** Using the voltage divider rule, determine the voltages V_{1}, and V_{3} for the series circuit of figure 1.58)

**Solution;**

V_{1 }= R_{1} E / R_{T} = (2 KΩ) (45 V)/ 2 KΩ + 5 KΩ + 8 KΩ = (2 KΩ) (45 V)/ 15 KΩ

= (2 x 10^{3} Ω) (45 V)/ 15 x 10^{3} Ω = 90 V / 15 = 6V

V3 = R2 E / RT = (8 KΩ) (45 V) / 15 KΩ = (8 x 10^{2} Ω) (45 V)/ 15 x 10^{3} Ω

= 360 V / 15 = 24 V

**Example 33;** Determine the value of R2 in figure 1.59 to establish a total resistance of 9 KΩ.

**Solution;**

R_{T} = R_{1} R_{2 }/ R_{1} + R_{2}

R_{T }(R_{1} + R_{2}) = R_{1} R_{2}

R_{T} R_{1 }+ R_{T} R_{2} = R_{1 }R_{2}

R_{T} R_{1} = R_{1} R_{2 }– R_{T} R_{2}

R_{T }R_{1} = (R_{1} -R_{T}) R_{2}

And R_{2} = R_{T} R_{1} / R_{1} – R_{T}

Substituting values;

R_{2 }= (9 KΩ) (12 KΩ) / 12 KΩ – 9 KΩ

= 108 KΩ/ 3 = 36 KΩ

**Example 34;** Determine the values of R_{1}, R_{2}, and R_{3} in figure 1.60, if R_{2} = 2R_{1} and R_{3} = 2R_{2 }and the total resistance is 16 KΩ.

Figure 1.60

**Solution;**

1 /R_{T} = 1 / R_{1 }+ 1/ R_{2} + 1 / R_{3}

1 / 16 KΩ = 1 / R_{1 }+ 1 / R_{2 }+ 1 / R_{3}

1 / 16 KΩ = 1 / R_{1} + 1 / 2R_{1} + 1 / 4 R_{1}

Since R_{3} = 2 R_{2} = 2 (2 R_{1}) = 4 R_{1}

And 1 / 16 KΩ = 1 /R_{1} + 1 /2 [1 / R_{1}] + ¼ [1 / R_{1}]

1 / 16 KΩ = 1.75 [1 / R_{1}]

With R_{1 }= 1.75 (16 Ω) = 28 KΩ

**Example 35;** For the parallel network of figure 1.61,

(a). Calculate R_{T}

(b). Determine I_{S}

(c). Calculate I_{1 }and I_{2 }and demonstrate that I_{S} = I_{1} + I_{2}

Figure 1.61

**Solution:**

(a). R_{T} = R_{1} R_{2} / R_{1} R_{2 }= (9 Ω) (18 Ω) / 9 Ω + 18 Ω = 162 Ω / 27 = 6 Ω

(b). I_{S} = E / R_{T} = 27 V / 6 Ω = 4.5 A

(c). I_{1} = V_{1} / R_{1 }= E / R_{1} = 27 V / 9 Ω = 3 A

I_{2} = V_{2} / R_{2} = E / R_{2} = 27 V / 18 Ω = 1.5 A

I_{S} = I_{1} + I_{2}

4.5 A = 3A + 1.5 A

4.5 A = 4.5 A (checks)

**Example 36;** Given the information provided in figure 1.62

(a). Determine R_{3}

(b). Calculate E

(c). Find I_{S}

(d). Find I_{2}

Figure 1.62

**Solution;**

1 / R_{T} = 1/ R_{1} + 1 / R_{2} + 1 / R_{3}

1 / 4 Ω = 1 / 10 Ω + 1 / 20 Ω + 1 / R_{3}

And 1 / R_{3} = 1/ 4 – 1 /10 – 1/ 20 = 5 -2 -1 / 20 = 2/ 20 or R_{3} = 20 /2 = 10 Ω

(b). E = V_{1 }= I_{1} R_{1} = (4A) (10 Ω) = 40 Ω

(c). I_{S} = E / R_{T} = 40 V/ 4 Ω = 10 A

(d). I_{2 }= V_{2} / R_{2 }= E / R_{2} = 40 V / 20 Ω =2 A

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