JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Current Electricity

When two identical batteries of internal resistance

1Ω

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

Ω

Answer: 4

Solution

Let each battery have emf

E

and internal resistance

r = 1Ω

.

Series connection:
Total emf =

2 E

, total internal resistance =

2 r = 2Ω

Current:

I_{1} = \frac{2 E}{R + 2}

Power in R:

J_{1} = I_{1}^{2} R = \frac{4 E ^{2} R}{( R + 2 ) ^{2}}

Parallel connection:
Total emf =

E

, total internal resistance =

\frac{r}{2} = 0.5Ω

Current:

I_{2} = \frac{E}{R + 0.5}

Power in R:

J_{2} = I_{2}^{2} R = \frac{E ^{2} R}{( R + 0.5 ) ^{2}}

Given:

J_{1} = 2.25 J_{2}

\frac{4 E ^{2} R}{( R + 2 ) ^{2}} = 2.25 \times \frac{E ^{2} R}{( R + 0.5 ) ^{2}}

\frac{4}{( R + 2 ) ^{2}} = \frac{2.25}{( R + 0.5 ) ^{2}}

\frac{( R + 0.5 ) ^{2}}{( R + 2 ) ^{2}} = \frac{2.25}{4} = \frac{9}{16}

\frac{R + 0.5}{R + 2} = \frac{3}{4}

4 (R + 0.5) = 3 (R + 2)

4 R + 2 = 3 R + 6

R = 4Ω

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer