Question

2. A set of resistors are connected to a 10.0 V battery as shown in the figure below. The capacitors have values of R1 = 300 2, R2 = 150 12, R3 = 250, R4 = 100 2 and Rs = 200 22. Determine the value of the following quantities (a) when the switch is open. • the current flowing through each resistor, • the potential drop across each resistor, and • the power dissipated in each resistor. (b) when the switch is closed and the capacitors are fully charged. • the current flowing through each resistor, • the potential drop across each resistor, and • the power dissipated in each resistor.

Answer 1

(a)

when switch is open, R3 and R2 will not contribute

R1 and R4 are in series

so,

R = R1 + R4 = 300 + 100 = 400 ohms

and this is in parallel with R5

so,

1/R_eq = 1/400 + 1/200

R_eq = 133.33 ohms

so,

I_eq = 10 / 133.3 = 0.075 A

so,

I (R5) = 0.05 A

I ( R1) = 0.025 A

I (R4) = 0.025 A

I ( R2) = I (R3) = 0 A

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V (R5) = 10 V

V (R4) = 2.5 V

V (R1) = 7.5 V

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P (R1) = 0.1875 W

P (R2) = 0

P (R3) = 0

P (R4) = 0.0625 W

P (R5) = 0.5 W

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(b)

Now, when switch is closed,

R4, R3, R2 are in parallel

so,

1/R = 48.38 ohms

this in in series with R1

so,

R = 348.38 ohms

finally , this is in parallel with R5

so,

R_eq = 127.1 ohms

Therefore,

I_eq = 0.0787 A

Therefore,

I (R5) = 0.05 A

I ( R1) = 0.0287 A

I (R4) = 0.013887 A

I ( R2) = 0.00926 A

I (R3) = 0.00555 A

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V (R5) = 10 V

V (R4) = V (R3) = V (R2) = 1.3887 V

V (R1) = 8.61 V

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P (R1) = 0.247 W

P (R2) = 0.0128 W

P (R3) = 0.0077 W

P (R4) = 0.019 W

P (R5) = 0.5  W