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In the circuit, the cells E1 and E2 have...

In the circuit, the cells `E_1 and E_2` have emfs of 4V and 8V and internal resistance `0.5 Omega and 1.0 Omega,` respectively. Calculate the current through `6 Omega` resistance.

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Net emf = 8-4 = 4 v, `R_(AB) = 6xx3//9 = 2Omega`

Total resistance of the circuit is `8Oemga.` So
`I_(1) = 4//8 = 0.5 A, V_(AB) = 0.5xx2 = 1V, I_(2) = 1//3A`, and
`I_(3) = 1//6A`
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In the circuit shown in fig. the cell has emf 10V and internal resistance 1 Omega

CENGAGE PHYSICS-ELECTRIC CURRENT AND CIRCUIT-Exercise 5.2
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