Question

5. Five resistors are connected to an ideal battery 8-24 V in a circuit as shown through; (b) power dissipation and (C) voltage across the 2.4 2 resistor? . What are the (a) current 6.00 Ω 240 Ω 8 6.00 Ω 6.00 Ω 9.00 Ω

Answer 1

here 6 and 9 are in parallel, so equivalent resistance becomes

R1 = 6*9 / ( 6+ 9) = 3.6 ohm

now 2.4 ohm is in series

R2 = 3.6 + 2.4 = 6

Now the middle 6 ohm resistor is in parallel

R3 = 6*6 / (6+6) = 3

now resistance 6 om connected with battery is in series

so net resistance becomes

R = 9 ohm.

net current through circuit

i = V/R = 24/9 = 2.67 A

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a)

Current through 2.4 ohm

i' = i ( 2.4 + 3.6)/( 6 + 2.4 + 3.6)

i' = 1.333 A

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b)

P = (i') ^2 * R = 1.333^2 * 2.4 = 4.267 W

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c)

V = 1.333* 2.4 = 3.1992 V

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