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A 3.0-ω resistor is connected in parallel with a 6.0-ω resistor. this combination is then connected in series with a 4.0-ω resistor. the resistors are connected across an ideal 12-volt battery. how much power is dissipated in the 3.0-ω resistor? a 3.0-ω resistor is connected in parallel with a 6.0-ω resistor. this combination is then connected in series with a 4.0-ω resistor. the resistors are connected across an ideal 12-volt battery. how much power is dissipated in the 3.0-ω resistor? 12 w 2.7 w 6.0 w 5.3 w

Respuesta :

first of all a 3 ohm resistance is connected in parallel with 6 ohm resistance so we will have

[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

[tex]\frac{1}{R} = \frac{1}{3} + \frac{1}{6}[/tex]

[tex]R = 2 ohm[/tex]

now it is connected in series with 4 ohm resistance

So we will have net resistance given by

[tex]R_{net} = R_1 + R_2[/tex]

[tex]R_{net} = 2 + 4 = 6 ohm[/tex]

Now this combination of 6 ohm resistance is connected across a battery of 12 V

so now we will have total current in the circuit calculated by ohm's law

[tex]V = i*R[/tex]

[tex]12 = i*6[/tex]

[tex]i = 2 A[/tex]

now this 2 A current will divide in 3 ohm and 6 ohm resistance in the ratio of 2:1

so current in 3 ohm resistance is given by

[tex]i = \frac{2}{2+1}*2= 1.33 A[/tex]

now power dissipated in 3 ohm resistance is given by

[tex]P = i^2 * R[/tex]

[tex]P = (1.33)^2* 3[/tex]

[tex]P = 5.33 Watt[/tex]

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