IIT-JEE 2009 · Paper 1 · Physics Q51

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PhysicsMultiple correct+4 / -1Question 51 of 60

For the circuit shown in the figure

A 24 V battery drives current I through R₁ = 2 kΩ in series with the parallel combination of R₂ = 6 kΩ and R_L = 1.5 kΩ.

the current

I

through the battery is 7.5 mA

the potential difference across

R_L

is 18 V

ratio of powers dissipated in

R_1

and

R_2

is 3

R_1

and

R_2

are interchanged, magnitude of the power dissipated in

R_L

will decrease by a factor of 9

Answer and worked solution

Official answer: A, D

Current. $R_2\parallel R_L=\dfrac{6\times1.5}{7.5}=1.2$ kΩ; total 3.2 kΩ; $I=\dfrac{24}{3.2}=7.5$ mA. (A) is true.

Voltage across $R_L$ . $7.5\times1.2=9$ V, not 18 V.

Powers. $P_1=I^2R_1=112.5$ mW and $P_2=\dfrac{9^2}6=13.5$ mW: ratio $\frac{25}3$ , not 3.

Interchanged. $R_1$ = 6 kΩ in series with $2\parallel1.5=\frac67$ kΩ: $V_L=24\times\dfrac{6/7}{6+6/7}=3$ V. Power in $R_L$ falls by $\left(\frac93\right)^2=9$ . (D) is true.

(A), (D)

Topics:Series and parallel resistors Electrical power and heating

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IIT-JEE 2009 · Paper 1, Physics question 51