Current Electricity

JEE Main level, single correct, medium

265 of 380

Question 265 of 380

Three resistors of 4Ω,6Ω4 \Omega, 6 \Omega and 12Ω12 \Omega are connected in parallel and the combination is connected in series with a 1.5 V1.5 \mathrm{~V} battery of 1Ω1 \Omega internal resistance. The rate of Joule heating in the 4Ω4 \Omega resistor is

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Answer: (C) 0.25 W0.25 \mathrm{~W}

Resistors 4Ω,6Ω4 \Omega, 6 \Omega and 12Ω12 \Omega are connected in parallel, its equivalent resistance (R)(R) is given by

1R=14+16+112⇒R=126=2Ω\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12} \Rightarrow R=\frac{12}{6}=2 \Omega
Again RR is connected to 1.5 V1.5 \mathrm{~V} battery whose internal resistance r=1Ωr=1 \Omega.
Equivalent resistance now,
R′=2Ω+1Ω=3ΩR^{\prime}=2 \Omega+1 \Omega=3 \Omega
Current, Itotal =VR′=1.53=12 AI_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \mathrm{~A}
Itotal =12=3x+2x+x=6x⇒x=112\begin{aligned} & I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x \\ & \Rightarrow x=\frac{1}{12} \end{aligned}
∴\therefore Current through 4Ω4 \Omega resistor =3x=3 x
=3×112=14 A=3 \times \frac{1}{12}=\frac{1}{4} \mathrm{~A}
Therefore, rate of Joule heating in the 4Ω4 \Omega resistor
=I2R=(14)2×4=14=0.25 W=I^2 R=\left(\frac{1}{4}\right)^2 \times 4=\frac{1}{4}=0.25 \mathrm{~W}

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