Problems on Combination of Resistors

Last Updated : 29 Sep, 2021

Resistors are the devices that offer resistance to the current. In our daily lives, our devices have more than one resistance in the circuits. It becomes essential to study the effect of different arrangements of resistances and their effects on the circuit. Often in real situations, it is required to calculate the required resistance for a full circuit or sometimes a portion of the circuit. In such cases, the knowledge of calculating equivalent resistances can be useful, let’s see these concepts in detail.

Resistors and Resistance

Resistors are electrical devices that restrict the flow of current in a circuit. It is an ohmic device, which means that it follows the ohms’ law V = IR. Most of the circuits have only one resistor, but sometimes more than one resistor can be present in the circuit. In that case, the current flowing through the circuit depends on the equivalent resistance of the combination. These combinations can be arbitrarily complex, but they can be divided into two basic types:

Series Combination
Parallel Combination

Series Combination

In the figure given below, three resistors are connected in series with the battery of voltage V. In these types of combinations, resistors are usually connected in a sequential manner one after another. The current through each resistor is the same. The figure on the right side shows the equivalent resistance of the three resistances. In the case of the series combination of resistances, the equivalent resistance is given by the algebraic sum of the individual resistances.

Let V₁, V₂, and V₃ be the voltages across all three resistances. It is known that the current flowing through them is the same.

V = V₁ + V₂ + V₃

Expanding the equation,

IR = IR₁ + IR₂ + IR₃

R = R₁ + R₂ + R₃

Parallel Combination

In the figure given below, three resistors are shown which are connected in parallel with a battery of voltage V. In this type of connection, the resistors are usually connected on parallel wires originating from a common point. In this case, the voltage through each resistor is the same. The figure on the right side shows the equivalent resistance of the three resistances.

The equivalent resistance of the given circuit is,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

In general for resistors R₁, R₂, R₃,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ....$

Sample Problems

Question 1: Three resistances of 6, 10, and 20 ohms are connected in series. Find the equivalent resistance for the system.

Answer:

The formula for series resistance is given by,

R = R₁ + R₂ + R₃

Given: R₁ = 6, R₂ = 10 and R₃ = 20

substituting these values in the equation,

R = R₁ + R₂ + R₃

⇒ R = 6 + 10 + 20

⇒ R = 36 Ω

Question 2: Three resistances of 1, 1, and 2 ohms are connected in parallel. Find the equivalent resistance for the system.

Answer:

The formula for parallel resistance is given by,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ....$

Given: R₁ = 1, R₂ = 1 and R₃ = 2

substituting these values in the equation,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ....$

⇒ $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

⇒ $\frac{1}{R} = \frac{2 + 2 + 1}{4}$

⇒ $R = \frac{4}{5}$ Ω

Question 3: Find the equivalent resistance for the system shown in the figure below.

Answer:

The formula for parallel resistance is given by,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ....$

and the formula for series resistance is given by,

R = R₁ + R₂ + R₃ + ….

This is combination of both parallel and series capacitances.

substituting these values in the equation,

R₁ = 10 μF ,R₂ = 2.5 μF

R= R₁ + R₂

⇒ R = 10 + 2.5

⇒ R = 12.5

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

⇒ $\frac{1}{R} = \frac{1}{12.5} + \frac{1}{0.3}$

⇒ $\frac{1}{R} = \frac{12.8}{(12.5)(0.3)}$

⇒ $R = 0.29$ Ω

Question 4: Find the equivalent resistance for the system shown in the figure below:

Answer:

The formula for parallel resistance is given by,

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ....$

and the formula for series resistance is given by,

R = R₁ + R₂ + R₃ + ….

This is combination of both parallel and series capacitances.

substituting these values in the equation,

R₁ = 100 μF ,R₂ = 25 μF

R= R₁ + R₂

⇒ R = 100 + 25

⇒ R = 125 Ω

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

⇒ $\frac{1}{R} = \frac{1}{125} + \frac{1}{3}$

⇒ $\frac{1}{R} = \frac{128}{(125)(03)}$

⇒ $R = 0.29$ Ω

Question 5: An electric heater is connected to a battery. How does the resistance of the circuit changes when another similar electric heater is added in series with the original one?

Resistance is doubled.
Resistance is halved.
Resistance remains constant.
Resistance is tripled.

Answer:

The component that is added to the circuit is in series. Let’s say resistance of the heater is R. It is known that in when connected series, the equivalent resistance is the algebraic sum of the individual resistances.

R_new = R + R

⇒ R_new = 2R

Thus, the resistance is doubled..

Answer (1).

Question 6: Find the equivalent resistance for the circuit given below.

Answer:

For such problems, break the circuit into small problems.

The resistances 4 and 2 are in parallel, calculating the equivalent resistance of this combination.

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

⇒ $\frac{1}{R} = \frac{1}{4} + \frac{1}{2}$

⇒ $\frac{1}{R} = \frac{3}{4}$

⇒ $R = 1.33$ Ω

The resistances 7 and 5 are in series, calculating the equivalent resistance of this,

R = R₁ + R₂

⇒ R = 7 + 5

⇒ R = 12 Ω

Now these two branches are in parallel, so we have a new combination of resistances of 12 and 1.33 which are in parallel.

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

⇒ $\frac{1}{R} = \frac{1}{12} + \frac{3}{4}$

⇒ $\frac{1}{R} = \frac{10}{12}$

⇒ R = 1.2 Ω