JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Current Electricity

For the circuit shown in the figure

(A)the current $I$ through the battery is 7.5 mA✓ Correct
(B)the potential difference across $R_{L}$ is 18 V
(C)ratio of powers dissipated in $R_{1}$ and $R_{2}$ is 3
(D)if $R_{1}$ and $R_{2}$ are interchanged, magnitude of the power dissipated in $R_{L}$ will decrease by a factor of 9✓ Correct

Answer: (A), (D)

Solution

Applying Kirchhoff's laws to the circuit with

R_{1} = 2

kΩ,

R_{2} = 6

kΩ,

R_{L} = 1.5

kΩ, and battery voltage 24 V. Let

I

be the current through the battery and

i

be the current through

R_{L}

. From the loop equations:

24 - 2000 I - 6000 (I - i) = 0

and

24 - 2000 I - 1500 i = 0

. Solving gives

I = 7.5

mA and

i = 6

mA. When

R_{1}

and

R_{2}

are interchanged, solving the new equations gives

I^{'} = 3.5

mA and

i^{'} = 2

mA. The power in

R_{L}

is proportional to

i^{2}

, so

\frac{P}{P ^{'}} = \frac{6 ^{2}}{2 ^{2}} = 9

. Therefore, options (A) and (D) are correct.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · One or more correct options