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A network of resistore is connected to a 16 V battery with internal resistance of 1Omega, as shown in (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor.(c ) obtain the voltage drops V _(AB), V _(BC) and V _(CD) |
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Answer» Solution :(a) The network is a simple series and parallel combination of resistors. First the two `4Omega` resistors in parallel are equivalent to a resistor `= [(4xx 4) //(4 +4) Omegta = 2 alpha` In the same way. The `12 Omega and 6 Omega` RESISTORES in parallel are equivalent ot a resistor of ` [ (12 xx 6) //(12+ 6) ]Omega = 4 Omega` The equivalent resistance R of the networn is obtained by conmbining these resistore `(2 Omega and 4 Omega)` with `1 Omega` in series. that is. `R = 2 Omega + 4 Omega + 1Omega = 7 Omega` (b)The total CURRENT In in the circuit is `I = (in )/(R +r) = (16 v)/((7 +1) Omega) = 2A` Consider the restors between A and B. If `I _(1)` is the current in one of the `4Omega` resistors and ` I (2)` the current in the other. `I _(1) xx 4 = I _(2) xx 4` then is, ` I _(1) = I _(2),` which is otherwise obvious from the symmetry of the two arma. But `I _(1) + I _(2) = I =2 A.` THUS. `I _(1) = I _(2) =1A` that , is current in each `4Omega` resistor is 1A. Current in `1Omega` resistor between B and C would be 2 A. Now, consider the resistance between C and D. If `I _(3)` is the current in the `12Omega` resistor, and `I_(4)` in the `6 Omega` RSISTOR. `I _(3) xx 12 =I _(4) xx 6,i.e., I _(4) = 2I _(3)` But, `I_(3) + I _(4) =I =2A` Thus, ` I _(3) = ((2)/(3)) A, I _(4) = ((4)/(3)) A` that is, the current in the `12Omega` resistor is (2/3) A, whithe the current in the `6 Omega` resistor is (4/3) is `V _(AB) = I _(A) xx 4= 1A xx 4 Omega = 4V.` This can also be obtained by multtplying the total current between A and B by the equivalent resistance between A and B, that is. `V_(AB) =2A xx 2 Omega =4V` The voltage drop across BC is `V _(BC)=2 A xx 1 Omega =2 V` Finally, the voltage drop across CD is `V _(CD) =12 Omega xx I _(3) =12 Omega xx ((2)/(3)) A =8V.` This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D. that is `V _(CD) =2 A xx 4 Omega =8V` Note that the total voltage drop across AD is `4 V + 2V +8V =14V.` Thus, the terminal voltage of the bettery is 14 V, while its emf is 16 V. The loss of the voltege `(=2V)` is accounted for by the internal resistance `1 Omega` of the bettery `[2 A xx 1 Omega = 2V].` |
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