Problem 2.24 Constants The currents i1 and i2 in the circuit...

Problem 2.24 Constants The currents i1 and i2 in the circuit in (Figure 1) are 28 A and 21 A, respectively. The source voltages are vg1 = 322 V and vg2 = 364 V. Part A Find the power supplied by each voltage source. Express your answers in watts using three significant figures separated by a comma. p322v, p364v = 1.13x10^4, 1.02x10^4 W Part B Find the total power dissipated in the resistors. Express your answer to three significant figures and include the appropriate units. pdis = 4.17 mW

Transcript

00:01 In this question, we have i1 is equal to 28 ampere, i2 equal to 21 ampere, vz1 is equal to 32 volt, and vz22 is equal to 364 volt.

00:13 The given circuit is shown in this figure.

00:18 This is loop ia, this is loop iv, and we can write this is ic.

00:31 In this question in part a we have to find power supplied by each voltage source power supplied by source vz1 and vg2.

00:44 And in part b we have to calculate total power dissipated in the register.

00:50 So first we are going to calculate power supplied by each voltage source.

00:58 From figure from the given circuit from circuit we can write i a minus i c equal to i 1 here i 1 is 28 so we can write this equation i a minus ic is equal to 28 equation number 1 now i b minus i c is equal to 28 equation number 1 now i b minus i c is equal to i2 here i 2 is 20 1 ampere so this equation is i b minus i c is equal to 21 this is equation number 2 now writing outer loop equation which is equal to minus v g 1 minus vg 2 plus 2 i a plus 80 plus 80 i c plus 2 i b equal to 0 now substituting values of vg 1 and vg 2 in this equation which is minus 3 22 22 minus 3 64 plus 2 i a plus 80 i c plus 80 i c plus 2 i c plus 2 i b equal to 0 now on solving this equation this equation becomes 2 i a 2 i a plus 80 ic 80 i c plus 2 i b equal to 6 86 now on solving this i a plus i b plus 40 i c equal to 3 43 this is equation number 3 now on solving these three equations 1 2 and 3 we find the value of i 1 i 2 and i 3 we find the value of i 1 i 2 and i 3 so on solving these three equations i 1, i2 and ic is given by from equation 1 to 3 from equation 1, equation 2 and equation 3.

04:20 We got values of i 1, i 2 and i 3.

04:24 I 1 is equal to 35 ampere, i 2 and i .3.

04:29 I 2 is equal to 28 ampere and i 3 is equal to 7 amper.

04:41 Now the power dissipated in vg1 is given by the power supplied in vg1.

05:02 This is given by pvg1 is equal to vg1, multiply ia on substituting values.

05:19 Vg1 and i in this equation which is 322 multiply 35 on solving this vg 1 is equal to 11 ,000 270 watt which is equal to p vg 1 power supplied by voltage vg 1 is 1 .13 multiplied 10 is power 4 watt.

05:59 This is power supplied by vg1.

06:02 Now calculate power supplied by vg2 which is given by power supplied in power supplied by vg2 which is equal to p vg2 equal to v g2 multiply ib ib which is equal to 364 multiply 30 364 ib is 28 therefore this equation is 10 ,1192 watt this equation is equal to p v g2 is equal to 1 .01 multiplied 10 inch power 4 watt.

07:27 This is power supplied in vg2.

07:35 Power supplied by those vg2 is 1 .01 multiplied standish to power 4...

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