what is the value of r? note that this new resistor is equivalent to the combination of

Resistors are the devices that offer resistance to the current. In our daily lives, our devices have more one resistance in the circuits. It becomes essential to study the effect of different arrangements of resistances and their effects on the circuit. Ofttimes 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, permit's run across these concepts in particular.

Resistors and Resistance

Resistors are electrical devices that restrict the menstruum of current in a circuit. It is an ohmic device, which ways that information technology follows the ohms' law 5 = IR. Almost of the circuits have only one resistor, but sometimes more than one resistor can exist 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 circuitous, but they can be divided into two basic types:

Series Combination
Parallel Combination

Series Combination

In the effigy given below, three resistors are connected in serial with the battery of voltage Five. In these types of combinations, resistors are usually connected in a sequential manner one afterwards 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 past the algebraic sum of the private resistances.

Let Five₁, Five₂, and V_three be the voltages across all three resistances. Information technology is known that the current flowing through them is the same.

V = V_one + Five_two + V₃

Expanding the equation,

IR = IR_one + IR₂ + IR_three

R = R_one + R_ii + R_three

Parallel Combination

In the figure given below, three resistors are shown which are connected in parallel with a battery of voltage V. In this blazon of connection, the resistors are usually connected on parallel wires originating from a common indicate. In this case, the voltage through each resistor is the aforementioned. The figure on the right side shows the equivalent resistance of the iii 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 full 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: 3 resistances of six, 10, and twenty ohms are connected in series. Find the equivalent resistance for the system.

Answer:

The formula for series resistance is given past,

R = R_ane + R₂ + R₃

Given: R₁ = 6, R₂ = 10 and R₃ = twenty

substituting these values in the equation,

R = R_one + R₂ + R₃

⇒ R = half dozen + ten + 20

⇒ R = 36 Ω

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

Answer:

The formula for parallel resistance is given past,

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

Given: R₁ = i, R_ii = ane and R_three = 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 three: Find the equivalent resistance for the system shown in the effigy 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 = R_i + R₂ + R₃ + ….

This is combination of both parallel and series capacitances.

substituting these values in the equation,

R_one = ten μF ,R₂ = ii.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 four: Find the equivalent resistance for the arrangement shown in the effigy beneath:

Reply:

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 five: An electrical heater is connected to a bombardment. How does the resistance of the circuit changes when another similar electrical 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 serial. Permit'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: Discover the equivalent resistance for the circuit given below.

Answer:

For such problems, intermission the excursion into small problems.

The resistances 4 and two 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 seven and 5 are in series, calculating the equivalent resistance of this,

R = R₁ + R₂

⇒ R = 7 + 5

⇒ R = 12 Ω

Now these 2 branches are in parallel, so we take 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.ii Ω