Question

Three resistors of $4.0\Omega $ , $6.0\Omega $ , and $10.0\Omega $ are connected in series. What is their equivalent resistance?

Answer 1

Hint: We must have to remember that there are two types of connections that can be made in the electric circuit: series and parallel. The equivalent resistance is calculated using two formulas – one is for series connection and another is for parallel connection. For the problem, use the formula of equivalent resistance of the resistors connected in series.

Formula used:
We must have to know that the equivalent resistance for three series resistors ${R_1}$ ,${R_2}$ and ${R_3}$ is, $R = {R_1} + {R_2} + {R_3}$

Complete answer:
As we know, the equivalent resistance of a branch of resistors connected in series is equal to the algebraic addition of the individual resistances.
The equivalent resistance for three series resistors ${R_1}$ ,${R_2}$ and ${R_3}$ is, $R = {R_1} + {R_2} + {R_3}$
Given that, ${R_1} = 4.0\Omega $ , ${R_2} = 6.0\Omega $ , and ${R_3} = 10.0\Omega $.
So, $R = 4 + 6 + 10 = 20\Omega $
Hence, the equivalent resistance for the three given resistors is, $R = 20\Omega $ (answer).

Note:
We must have to know that the resistance is used in a circuit to control the current passing through the circuit.
We have to know that more resistance implies that less current is flowing through the circuit. The equivalent resistance is another way of indicating 'total' resistance, which we evaluate differently for series and parallel circuits. In a series circuit, the different elements are connected in a single, continuous loop.
Since the current through each element is the same, the equality is simplified to an equivalent resistance, that is just the addition of the resistances of the individual resistors.
In a parallel circuit, the potential across each of the elements is the same, and the total current is the addition of the currents passing through each component.