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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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A 5V battery with internal resistance $2\Omega$ and a 2V battery with internal resistance $1\Omega$ are connected to a $ 10 \Omega $ resistor as shown in the figure. The current in the $ 10\Omega $ resistor is

 

$\begin {array} {1 1} (A)\;0.27\: A\: P_2 \: to \: P_1 & \quad (B)\;0.03\: A\: P_1 \: to \: P_2 \\ (C)\;0.03\: A\: P_2 \: to \: P_1 & \quad (D)\;0.27\: A\: P_1 \: to \: P_2 \end {array}$

 

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1 Answer

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$V_{p_2}-V_{p_1} = \large\frac{\large\frac{5}{2}+\large\frac{0}{10}-\large\frac{2}{1}}{\large\frac{1}{2}+\large\frac{1}{10}+\large\frac{1}{1}}$
$ I = \large\frac{V_{p_2}-V_{p_1}}{10} $ $0.03 $ from $ p_2 \rightarrow p_1$
Ans : (c)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 14, 2014 by balaji.thirumalai
 

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