Answer

Verified

418.5k+ views

**Hint**

To find the current in the circuit we need to find the equivalent resistance that is present in the circuit. Then by using Ohm's law with the given cell of e.m.f. of 2V and the calculated equivalent resistance we can find the current in the circuit.

Formula Used: In this solution, we will be using the following formula,

${R_{eq}} = {R_1} + {R_2} + {R_3} + ....$ where ${R_{eq}}$ is the equivalent resistance when the resistances are placed in series.

and $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....$ where ${R_{eq}}$ is the equivalent resistance when the resistances are placed in a parallel circuit.

And from Ohm’s law,

$V = IR$ where $V$ is the potential of the cell and $R$ is the equivalent resistance.

**Complete step by step answer**

To solve the given problem we need to first find the equivalent resistances in the given circuit. Now in the circuit, we name the points as,

So in between the points A and B we can simplify the circuit as,

So on the top wire 2 resistances are in series, so the equivalent resistance is given by the formula

${R_{eq}} = {R_1} + {R_2} + {R_3} + ....$

Therefore, ${R_{eq}} = \left( {30 + 30} \right)\Omega $

$ \Rightarrow {R_{eq}} = 60\Omega $

Now the resistance in the top and bottom wire are in parallel, so we get the equivalent resistance from the formula,

$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....$

So substituting ${R_1} = {R_{eq}} = 60\Omega $ and ${R_2} = 30\Omega $ we get

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

$ \Rightarrow \dfrac{1}{R} = \dfrac{1}{{60}} + \dfrac{1}{{30}}$

On doing LCM of the denominator,

$\dfrac{1}{R} = \dfrac{{1 + 2}}{{60}} = \dfrac{3}{{60}}$

So on taking the reciprocal we get the value of the equivalent resistance as

$R = \dfrac{{60}}{3} = 20\Omega $

So now in the given circuit, the resistance is $R = 20\Omega $ and the potential across the cell is given in the figure, $V = 2V$. So from Ohm’s law, the current in the circuit is,

$i = \dfrac{V}{R}$

So substituting the values we get

$i = \dfrac{2}{{20}}$

Cancelling 2 from numerator and denominator we have,

$i = \dfrac{1}{{10}}A$

**Therefore the correct answer is C.**

**Note**

In the given circuit the resistances are in parallel. So the current passing gets divided into the two streams and the current in a particular stream is found by dividing the potential drop of the resistance by the resistance value. But the potential across the resistances in the two wires will be the same, that is the potential drop across the cell 2V.

Recently Updated Pages

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

Trending doubts

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

10 examples of evaporation in daily life with explanations

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference Between Plant Cell and Animal Cell

What are the monomers and polymers of carbohydrate class 12 chemistry CBSE