A network of resistances is connected to a 16 V battery with internal resistance of `1 Omega`, as shown in Fig. 4.33. (a) Compute epuivalent resistance of the network, (b) obtain the current In in each resistor, and (c ) obtain the voltage drops `V_(AB), V_(BC) " and " V_(CD)`.

A network of resistances is connected to a 16 V battery with internal

1.	A network of resistances is connected to a 16 V battery with internal resistance of `1 Omega`, as shown in Fig. 4.33. (a) Compute epuivalent resistance of the network, (b) obtain the current In in each resistor, and (c ) obtain the voltage drops `V_(AB), V_(BC) " and " V_(CD)`.
Answer» (a) The network is a simple series and parallel combination of resistors. First the two `4 Omega` resistors in parallel are equivalent to a resistor `= [(4 xx 4)//(4 + 4)] Omega = 2 Omega`. In the same way, the `12 Omega and 6 Omega` resistors in parallel are equivalent to a resistor of `[(12 xx 6)//(12 + 6)] Omega = 4 Omega`. The equivalent resistance R of the network is obtained by combining these resistors `(2 Omega and 4 Omega)` with `1 Omega` in series, that is, `R = 2 Omega + 4 Omega + 1 Omega = 7 Omega` (b) The total current `I` in the circuit is `I = (epsilon)/(R+r)=(16 V)/((7+1)Omega)=2A` Consider the resistors between A and B. If `I_(2)` is the current in one of the `4 Omega` resistors and `I_(2)` the current in the other `I_(1) xx 4 = I_(2)xx4` that is, `I_(1)=I_(2)`, which is otherwise obvious from the symmetry of the two arms. But `I_(1)+I_(2)=I=2A`. Thus, `I_(1)=I_(2)=1A` that is, current in each `4 Omega` resistor is 1A. current in `1 Omega` resistor between B and C would be 2A Now, consider the resistance between C and D. If `I_(3)` is the current in the `12 Omega` resistor, and `I_(4)` in the `6 Omega` resistor, `I_(3)xx12=I_(4)xx6` 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 `12 Omega` resistor is `(2//3)A`, while the current in the `6 Omega` resistor is (4/3)A The voltage drop across AB is `V_(AB)=I_(1)xx4=1 A xx 4 Omega = 4V` This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is, `V_(AB) = 2 A xx 2 Omega = 4V` The voltage drop across BC is `V_(BC) = 2A xx 1 Omega = 2V` 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 = 8 V` Note that the total voltage drop across AD is `4 V + 2 V + 8 V = 14 V`. Thus, the terminal voltage of the battery is 14 V, while its emf is 16 V. The loss of the voltage (= 2 V) is accounted for by the internal resistance `1 Omega` of the battery `[2 A xx 1 Omega= 2 V]`.

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