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JEE Main 2014 (11 Apr, Shift 1) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2014 (11 Apr, Shift 1)

Q183JEE Main 2014 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively:

Figure

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Answer: (c)

  1. Left rails: top 5050 V, bottom 00. Right rails: top aa, bottom bb with a−b=30a - b = 30 V.
  2. The two 5 Ω5\ \Omega carry the same current: 50−a5=b5⇒a+b=50\frac{50 - a}{5} = \frac{b}{5} \Rightarrow a + b = 50, so a=40a = 40, b=10b = 10, current =2= 2 A.
  3. 10 Ω10\ \Omega: 3010=3\frac{30}{10} = 3 A, so the 3030 V battery gives 3−2=13 - 2 = 1 A.
  4. 20 Ω20\ \Omega: 5020=2.5\frac{50}{20} = 2.5 A, so the 5050 V battery gives 2.5+2=4.52.5 + 2 = 4.5 A.
  5. So the answer is (c) 4.5 and 1.
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