Please I need help with this question The circuit shown in the figure below is c

Question

Please I need help with this question

The circuit shown in the figure below is connected for 2.30 min. (Assume R 6.70 n, R2 2.50 n, and V 18.0 w.) 5.000 3.000 1.00 (2 4.00 V (a) Determine the current in each branch of the circuit. branch magnitude (A) direction E left branch middle branch right branch (b) Find the energy delivered by each battery 4.00 V battery 18.0 V battery (c) Find the energy delivered to each resistor. 6.70 n resistor 5.00 n resistor 1.00 n resistor 3.00 n resistor 2.50 resistor

Explanation / Answer

I = I1 + I2

-4 + I(6.70) + 6 ( I1) = 0---------left loop

-18 +4 + 6 (-I1 ) + I2 ( 5.50) =0-------right loop

solving loops

-4 + I1 ( 12.70 ) + I2 (6.70) = 0---(eq 2)

-14 - 6(I1) + 5.50 (I2) = 0---------(eq 3)

multiply eq ( 2) with 6 and multiply eq( 3) with 12.70

-24 + I1 (76.2) + 40.3 (I2) =0

-177.8 - I1 (76.2) + 69.85 (I2) =0

adding both the above equations,

-201.8 =- 110.15 (I2)

I2= 1.83 A

I1= - 0.653 A

I = 1.176 A

Energy delivered by 4v = 2.30 (60) ( 1.176^2 [6.70] + 0.653 ^ 2 [6]) = 1631.77 J

Energy deliverecby 18 V = 2.30 (60 ) [ 1.83^2 {6.50 } + 0.653 ^ 2 [6] ) = 3.3569 kJ apprx

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