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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Current Electricity

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are $R_1\: and \: R_2 (R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero then,

$\begin {array} {1 1} (A)\;R = R_2 \times (R_1+R_2)/R_2-R_1) & \quad (B)\;R=R_2-R_1 \\ (C)\;R=R_1R_2/(R_1+R_2) & \quad (D)\;R=R_1R_2/(R_2-R_1) \end {array}$

 

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1 Answer

Total current in the circuit is given by
$ I = \large\frac{2E}{R_1+R_2+R}$
since the potential difference across the sourse having internal resistance $R_2$ is zero
$E-R_2I=0$
$ \Rightarrow R = R_2-R_1$
Ans : (B)

 

answered Feb 15, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham
 

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