2.22 The current io in the circuit in Fig. P2.22 is 2 A. a)...

2.22 The current i_o in the circuit in Fig. P2.22 is 2 A. a) Find i_1. b) Find the power dissipated in each resistor. c) Verify that the total power dissipated in the circuit equals the power developed by the 180 V source. Figure P2.22 60 ?? 5 ?? 25 ?? 180 V 70 ?? 35 ??

Transcript

00:01 In the current question, we have been given the following circuit, resistance of 60 ohm, which is attached to 2 resistance of 5 ohm and 25 ohm and here there is a cell of 180 volt and there is a 70 ohm resistance here.

00:36 So, this is the circuit, it has been given that this is the current i0 which is 2 ampere and a current i1 is flowing through this, that if you label it as a, b, c, d, e, f, g and h.

01:03 I1 is flowing through h, e, then we have to find what is the current i1.

01:15 So, if we say that using the kirchhoff s current rule, this is i0 and since this is i1, we have i2 and this is i3, so we have i4 here and that is it.

01:44 So, from the kirchhoff s current rule, we have i0 minus i2 minus i3 is 0, this is equation number 1, then i4 plus i2 minus i1 is equal to 0, this is equation number 2 and from the kirchhoff s voltage rule by going in a clockwise direction in all the three sections, we have 60 times i0 plus 25 i2 minus 5 i4 is equal to 0, this is equation number 3, then minus 25 i2 plus 35 i3 minus 70 i1 is equal to 0, this is equation number 4 and the last equation is 5 i4 plus 70 i1 minus 180 is equal to 0.

03:04 So, we see that we have i0, i1, i2, i3, i4, we have 5 currents, i0 is known, so we have 5 equations.

03:15 So, we can set up a matrix equation of the following form, so the matrix will be this 5 cross 5 with values 1, so i0, i0 in the first equation is 0, so in the first equation i0 is 1, we have 1 0 minus 1 minus 1 0 0 minus 1 1 0 1 60 0 25 0 minus 5 0 minus 70 minus 25 35 0 and 0 70 0 0 and 5.

04:24 So, this is multiplied with i0, i1, i2, i3, i4 and the right hand side will be 0 0 0 0 180.

04:42 So, this is the matrix equation we have for all the 6 currents.

04:47 So, the only simplification that we will do over here is we will say if so this is the rho 1, let us say this is rho 0, rho 1, rho 2, rho 3 and rho 4, so we will say rho 4 is equal to rho 2 plus rho 3 plus rho 4, this is the rho simplification or rho modification that we will do, rho algebra.

05:20 So, we will get 1 0 minus 1 minus 1 0 0 minus 1 1 0 1 60 0 25 0 minus 5 0 minus 70 minus 25 35 0 and then finally we will get 60 plus 0 is 60 0 plus 70 0 minus 70 plus 70 that will be 0, 0 minus 25 plus 25 that is 0, 0 plus 35 plus 0 is 35 and 0 minus 5 plus 5 is 0.

06:13 So, this will be i0, i1, i2, i3, i4 and that will be equal to, the right hand side will remain the same because r2 and r3 has 0s in them.

06:31 So, we get the following equations from this, one is as we had got earlier i0 is equal to i2 plus i3 and i1 is equal to i2 plus i4...

Question

2.22 The current i_o in the circuit in Fig. P2.22 is 2 A. a) Find i_1. b) Find the power dissipated in each resistor. c) Verify that the total power dissipated in the circuit equals the power developed by the 180 V source. Figure P2.22 60 ?? 5 ?? 25 ?? 180 V 70 ?? 35 ??

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