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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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In the circuit shown, the current passing through the:

$\begin {array} {1 1} (A)\;3 \: \Omega \: resistor\: is\: 0.50\: A & \quad (B)\;3 \: \Omega \: resistor\: is\: 0.25\: A \\ (C)\;4 \: \Omega \: resistor\: is\: 0.50\: A & \quad (D)\;4 \: \Omega \: resistor\: is\: 0.20\: A \end {array}$

 

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1 Answer

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So the current through the 9V battery is 1A.
Also note, that the current through the 8 $ \Omega$ branch nearer to the battery,
is 0.5A and through the second 8 $ \Omega$ branch is $ 0.5 \times 0.5 = 0.25A.$
Hence, the current through the 4$ \Omega$ resistor is 0.25A.
Ans : (D)
answered Feb 3, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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