Question

Find the current drawn from the battery by a network of four resistors shown in the figure?

Answer 1

Hint: Resistance is a property of conductor due to which it resists the flow of electric current through it. A component that is used to resist the flow of electric current in a circuit is called a resistor. There are two ways to connect resistors- parallel and series.

Complete step by step solution:
To solve this question, first we need to understand the concept of combination of resistance. Resistors can be combined in 2 ways:-
(A) Resistors in Series: A series circuit is a circuit in which resistors are arranged in a single chain, resulting in common current flowing through them. The characteristic of this type of combination is that the current through all the resistors are the same but the voltage of each resistor is different.
(B) Resistors in Parallel: A parallel circuit is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together. The characteristic of this type of combination is that the voltage through all the resistors are the same but the current of each resistor is different.
Given,
 $ V = 3V $
 $ {R_1} = {R_2} = {R_3} = {R_4} = 10\Omega $
Here, $ {R_1} $ and $ {R_2} $ are connected in series
Also, $ {R_3} $ and $ {R_4} $ are in series
So we can say that,
 $ {({R_{eq}})_1} = {R_1} + {R_2} $
 $ {({R_{eq}})_1} = 10 + 10 $
 $ {({R_{eq}})_1} = 20\Omega .......(1) $
Similarly, $ {({R_{eq}})_2} = {R_3} + {R_4} $
 $ {({R_{eq}})_2} = 10 + 10 $
 $ {({R_{eq}})_2} = 20\Omega .......(2) $
Now, $ {({R_{eq}})_1} $ and $ {({R_{eq}})_2} $ are in parallel,
 $ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{{({R_{eq}})}_1}}} + \dfrac{1}{{{{({R_{eq}})}_2}}} $
Now by putting the values from equation (1) and (2), we get,
 $ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{20}} + \dfrac{1}{{20}} $
 $ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{10}} $
On taking the reciprocal on both the sides,
 $ {R_{eq}} = 10\Omega .......(3) $
From ohm’s law
 $ V = IR $
 $ I = \dfrac{V}{R} $
On putting the values of $ V $ and $ R $ , we get,
 $ I = \dfrac{3}{{10}}A $
 $ I = 0.3A $
Thus, Current drawn by the resistor from the battery is $ 0.3A $

Note:
In a parallel circuit, the net resistance decreases as more components are added, because there are more paths for the current to pass through whereas in a series circuit, the net resistance increases as more components are added.