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In the circuit shown in figure. Find the currents I,I_(1),I_(2) and I_(3) given that emf of th battery=2V, internal resistance of the battery =2Omega and resistance of the galvanomter =4Omega.

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Answer :`I=(47)/(91)A,I_(1)=(17)/(91)A,I_(2)=(30)/(91)A,I_(3)=-(1)/(91)A`


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