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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Current Electricity

BITSAT 2013 · Physics · Current Electricity

In the circuit shown in figure potential difference between points...

In the circuit shown in figure potential difference between points AA and BB is 16V16 \mathrm{ V}. the current passing through 2Ω2 \Omega resistance will be In the circuit shown in figure potential difference between points A and B is 16 V . the current passing through 2

Options

  1. A2.5A2.5 \mathrm{ A}
  2. B3.5A3.5 \mathrm{ A}
  3. C4.0A4.0 \mathrm{ A}
  4. Dzero

Correct answer

B. 3.5A3.5 \mathrm{ A}

Step-by-step solution

∴4i1+2(i1+i2)−3+4i1=16 V\therefore 4 \mathrm{i}_{1}+2\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)-3+4 \mathrm{i}_{1}=16 \mathrm{~V} Using Kirchhoff's second law in the closed loop we have 9-i_{2}-2 (i1+i2)=0Solvingequations(i)and(ii),wegeti1=\left(i_{1}+i_{2}\right)=0 Solving equations (i) and (ii), we get \mathrm{i}_{1}=1.5  A\mathrm{~A} and i2=\mathrm{i}_{2}=2  A\mathrm{~A}\thereforecurrentthroughcurrent through2  W\mathrm{~W} resistor =2+1.5=3.5 A=2+1.5=3.5 \mathrm{~A}
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