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ICSE Class 10 Physics 2020 Paper

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The diagram above shows three resistors connected across a cell of e.m.f. 1.8V1.8 V and internal resistance rr. Calculate:
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Question : 52 of 67
Marks: +1, -0
Current through 3 Ω3\,\Omega resistor.
Solution:  
Let the current through 3 Ω3\,\Omega resistor be ' ii '
A. So the current through 1.5 Ω1.5\,\Omega resistor will be (0.3−i)(0.3-i) A. As 3 Ω3\,\Omega resistor and 1.5 Ω1.5\,\Omega resistor are in parallel. So,
  3×i  =1.5(0.3−i)\;3 \times i\;=1.5(0.3-i)
⇒  3i  =0.45−1.5i\Rightarrow\;3 i\;=0.45-1.5 i
⇒  4.5i  =0.45\Rightarrow\;4.5 i\;=0.45
=  0.454.5=0.1  A=\;\frac{0.45}{4.5}=0.1\;A
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