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QJEEAdv22Ph2429: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph2429 Solution: The figure shows a circuit having eight resistances of each,…

The figure shows a circuit having eight resistances of 1Ω1 \Omega each, labelled R1R_{1} to R8R_{8} , and two ideal batteries with voltages ε1=12 V\varepsilon_{1}=12 \mathrm{~V} and ε2=6 V\varepsilon_{2}=6 \mathrm{~V} .

JEE Advanced 2022 Paper 1 Online Physics - Current Electricity Question 6 English

Which of the following statement(s) is(are) correct?

Multiple Choice Options

  • A The magnitude of current flowing through R1R_{1} is 7.2 A7.2 \mathrm{~A} .
  • B The magnitude of current flowing through R2R_{2} is 1.2 A1.2 \mathrm{~A} .
  • C The magnitude of current flowing through R3R_{3} is 4.8 A4.8 \mathrm{~A} .
  • D The magnitude of current flowing through R5R_{5} is 2.4 A2.4 \mathrm{~A} .

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 1 Online Physics - Current Electricity Question 6 English Explanation 1

Point A and B are at same potential so they can be merged/folded.

JEE Advanced 2022 Paper 1 Online Physics - Current Electricity Question 6 English Explanation 2

For loop 1


amp;−12x−12(x+y)−x+6=0amp;⇒−2x−y2=−6amp;⇒−4x−y=−12amp;⇒4x+y=12.......(1)\begin{aligned}& -\frac{1}{2} x-\frac{1}{2}(x+y)-x+6=0 \\\\&\Rightarrow -2 x-\frac{y}{2}=-6 \\\\&\Rightarrow -4 x-y=-12 \\\\&\Rightarrow 4 x+y=12 .......(1)\end{aligned}

For loop 2

amp;12y+1y+12+12(x+y)=0amp;⇒32y+y2+x2=−12amp;⇒2y+x2=−12amp;⇒4y+x=−24⇒4y+x−16x−4y=−24−48amp;⇒−15x=−72\begin{aligned}& \frac{1}{2} y+1 y+12+\frac{1}{2}(x+y)=0 \\\\&\Rightarrow \frac{3}{2} y+\frac{y}{2}+\frac{x}{2}=-12 \\\\&\Rightarrow 2 y+\frac{x}{2}=-12 \\\\& \begin{array}{l}\Rightarrow4 y+x=-24 \\\\\Rightarrow4 y+x-16 x-4 y=-24-48\end{array} \\\\&\Rightarrow -15 x=-72\end{aligned}

⇒\Rightarrow x=7215x=\frac{72}{15}

⇒\Rightarrow 4(7215)+y=124\left(\frac{72}{15}\right)+y=12

⇒yamp;=12−28815amp;=180−28815amp;=−10815=−7.2 A\begin{aligned}\Rightarrow y & =12-\frac{288}{15} \\\\& =\frac{180-288}{15} \\\\& =\frac{-108}{15}=-7.2 \mathrm{~A}\end{aligned}

Current in R1=7.2 AR_1=7.2 \mathrm{~A}

Current in R2=x+y2=(7215−10815)12R_2=\frac{x+y}{2}=\left(\frac{72}{15}-\frac{108}{15}\right) \frac{1}{2}

amp;=12(3615)=2.42 Aamp;=1.2 A\begin{aligned}& =\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2} \mathrm{~A} \\\\& =1.2 \mathrm{~A}\end{aligned}

Current in R3=x=4.8 AR_3=x=4.8 \mathrm{~A}

Current in R5=12x=2.4 AR_5=\frac{1}{2} x=2.4 \mathrm{~A}

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