Question

The total current supplied to the circuit by the battery is

A. 1A
B. 2A
C. 4A
D. 6A

Answer 1

Hint:The above circuit diagram can be simplified and presented in a way where the resistances appear in parallel and serial order with each other. The equivalent resistance in the whole circuit can be obtained by solving individually for series resistances and parallel resistances. Then by using Ohm’s Law, the total current supplied to the circuit can be found out.

Formula Used:
If resistances $$R_a$$ and $$R_b$$are connected in series, then the equivalent resistance is given by: $$R_{series}=R_a+R_b$$
If resistances $$R_a$$ and $$R_b$$ are connected in parallel, then the equivalent resistance is given by: $${\displaystyle \frac{1}{R_{parallel}}}={\displaystyle \frac{1}{R_a}}+{\displaystyle \frac{1}{R_b}}$$
Ohm’s Law is given by: $$V=IR$$
where, $$V$$ is the voltage, $$R$$ is the resistance and $$I$$is the current.

Complete step by step answer:
The above diagram can be simplified as given in the following diagram:

As in the simplified diagram, the resistances $$2\mathrm{\Omega }$$ and $$6\mathrm{\Omega }$$ are connected in parallel. So, the equivalent resistance between them will be
$${\displaystyle \frac{1}{R}}={\displaystyle \frac{1}{2\mathrm{\Omega }}}+{\displaystyle \frac{1}{6\mathrm{\Omega }}}={\displaystyle \frac{3\mathrm{\Omega }+1\mathrm{\Omega }}{6\mathrm{\Omega }}}={\displaystyle \frac{4}{6}}\mathrm{\Omega }={\displaystyle \frac{2}{3}}\mathrm{\Omega }$$
Therefore, $$R={\displaystyle \frac{3}{2}}\mathrm{\Omega }=1.5\mathrm{\Omega }$$

Then, the diagram can be again simplified as follows.

Now, all the resistances are in series. Then, the equivalent resistance in the circuit will be
$$R_{eq}=1.5\mathrm{\Omega }+1.5\mathrm{\Omega }+3\mathrm{\Omega }=6\mathrm{\Omega }$$
Therefore, the total resistance in the circuit is $$6\mathrm{\Omega }$$.
The voltage of the cell is given as $$6V$$.Thus, $$V=6V$$. The net current can be found out by Ohm’s Law
$$V=IR$$
where, $$V$$ is the voltage, $$R$$ is the total resistance in the circuit and $$I$$ is the current flowing through the circuit.
Here, $$V=6V$$and $$R=6\mathrm{\Omega }$$.
Thus, $$I={\displaystyle \frac{V}{R}}={\displaystyle \frac{6V}{6\mathrm{\Omega }}}=1A$$

Hence, option A is the correct answer.

Note: The battery is a combination of cells. An electric cell works as a source of emf. The emf generated is also known as voltage. Every cell generates a particular voltage and can supply a finite amount of current. Usually voltage and current of one cell is not enough for many practical applications. Therefore, they are always connected as a combination to form a battery.