# When two identical batteries of internal resistance $1\Omega $ each are connected in series across a resistor R, the rate of heat produced in R is ${{J}_{1}}$. When the same batteries are connected in parallel across R, the rate is ${{J}_{2}}$. If ${{J}_{1}}=2.25{{J}_{2}}$ then the value of R in $\Omega $ is,

Answer

Verified

279.3k+ views

**Hint:**When we connect two identical batteries in series the effective resistance in series combination will be the sum of those two internal resistances and in while the batteries are connected in parallel the reciprocal of those two internal resistances are added together to find the effective internal resistances. Then by calculating ${{I}_{1}}$and ${{I}_{2}}$ , we can find the heat in those two cases. Hence by substituting in the given equation we will get the value of R.

**Complete step by step answer:**

Given that the internal resistance $1\Omega $ is connected in series with a resistor R.

Then the total resistance in series combination becomes 2r+R.

Hence the current flow through ${{I}_{1}}$ is given by,

Current ${{I}_{1}}=\left[ \frac{2E}{2r+R} \right]$

Heat in the first case can be calculated by using the equation,

${{J}_{1}}=I_{1}^{2}Rt$

$\Rightarrow {{J}_{1}}={{\left( \frac{2E}{2r+R} \right)}^{2}}Rt$

Similarly we can calculate the current ${{I}_{2}}$. Here the same battery is connected in parallel across R. Thus,

${{I}_{2}}=\left[ \frac{E}{\frac{r}{2}+R} \right]$

Hence heat produced in the second case,

${{J}_{2}}=I_{2}^{2}Rt$

$\Rightarrow {{J}_{2}}={{\left( \frac{E}{\frac{r}{2}+R} \right)}^{2}}Rt$

Given that,

${{J}_{1}}=2.25{{J}_{2}}$

Substituting the values of ${{J}_{1}}$and ${{J}_{2}}$ in the above equation we get, ${{\left( \frac{2E}{2r+R} \right)}^{2}}Rt=2.25{{\left( \frac{E}{\frac{r}{2}+R} \right)}^{2}}Rt$

${{\left( \frac{2E}{2r+R} \right)}^{2}}Rt=2.25{{\left( \frac{2E}{r+2R} \right)}^{2}}Rt$

Rearranging the equation and cancelling the common terms we get,

${{\left( r+2R \right)}^{2}}=2.25{{\left( 2r+R \right)}^{2}}$

${{\left( r+2R \right)}^{2}}=\frac{9}{4}\times {{\left( 2r+R \right)}^{2}}$

$4{{\left( r+2R \right)}^{2}}=9{{\left( 2r+R \right)}^{2}}$

Taking the square root the above equation becomes,

$2\left( r+2R \right)=3\left( 2r+R \right)$

$R=4r$

Given that,

$r=1\Omega $

Then the value of R in $\Omega $ is,

$R=4\Omega $

**Note:**In series the effective resistance in series combination will be the sum of those two resistances and parallel the reciprocal of those two internal resistances are added together to find the effective internal resistances.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Define absolute refractive index of a medium

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Trending doubts

What is 1 divided by 0 class 8 maths CBSE

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

What is the past tense of read class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

How many crores make 10 million class 7 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE