c) Find the energy stored in cach capacitor. Ans. Fig. 4 Fig. 5 Fig. 6 C2 4 R1 1

Question

c) Find the energy stored in cach capacitor. Ans. Fig. 4 Fig. 5 Fig. 6 C2 4 R1 10 R4 C, 5 V R6 R2 4 2 RS 4 C3 R7 R3 8 2 9 V P4 - Ch20 (25/100) [Current, Ohm's law, electric power/energyl a) In Fig. 5, E = 9 V, r = 0.5 ohm, and Road = 9.5 ohm. Find the current, 1-7 Ans. b) Find the voltage on Ricad, Ans ;c) Find the power on Rioad, Ans. P5 - Ch21 (25/100) [Battery, load, resistor network] a) c) In Fig. 6, what is the current in R1? Ans.;b) the current in R2? Ans. The Current in R4? Ans ; d) the current in R5? Ans. ; e) the current in R6? Ans.

Explanation / Answer

P4)

(a)

I = E/(R + r) = 9/(9.5+0.5) = 9/10 = 0.9 A

(b)

voltage across R , VR = I*R = 0.9*9.5 = 8.55 V

power on R , PR = I^2*R = 0.9^2*9.5 = 7.695 W

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P5)

a)

R6 , R7 are in series R67 = 2 + 2 = 4

R5 , R67 are in parallel R567 = (R5*R67)/(R5+R67)

R567 = (4*4)/(4+4) = 2

R567 and R4 are in series R4567 = 10 + 2 = 12

R2 , R3 are in series R23 = R2 + R3 = 4 + 8 = 12 ohm

R4567 and R23 are in in parallel R234567 = (12*12)/(12+12) = 6

R234567 and R1 are in series

R1234567 = R1 + R234567 = 4 + 6 = 10 ohms

I1 = E/Rtototal = 5/10 = 0.5 A <<<--------ANSWER

(b)

voltage across R23, V23 = E - I1*R1 = 5 - (4*0.5) = 3

(b)

current in R2 = I2 = V23/R23 = 3/12 = 0.25 A

(c)

current in R4 = I4 = I1 - I2 = 0.5-0.25 = 0.25 A

(d)

voltage across R5 , V5 = I4*R567 = 0.25*2 = 0.5

current in R5 = I5 = V5/R5 = 0.5/4 = 0.125 A

(e)

current in R6 = V67/R67 = 0.5/4 = 0.125 A

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