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Three resistors 1Omega , 2Omega " and " ...

Three resistors `1Omega , 2Omega " and " 3Omega ` are combined in series. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Text Solution

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(a) Here `R_1 = 1Omega , R_2 = 2 Omega` and `R_3 = 3Omega `
Total resistance in series combination `R = R_1 + R_2 + R_3 = 1 + 2 + 3 = 6Omega`
(b) As `epsi = 12 V, r = 0 `, hence current
`I = (epsi)/(R+r ) = (12)/(6+0) = 12/6 = 2.0 A`
Potential drop across the three resistors are respectively
`V_1 = IR_1 = 2 xx 1 = 2V, V_2 = IR_2 = 2 xx 2 = 4V " and " V_3 = IR_3 = 2 xx 3 = 6V`
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