Question

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

Answer 1

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

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{3} + \frac{1}{6}$

$R = 2 ohm$

now it is connected in series with 4 ohm resistance

So we will have net resistance given by

$R_{net} = R_1 + R_2$

$R_{net} = 2 + 4 = 6 ohm$

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

$V = i*R$

$12 = i*6$

$i = 2 A$

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

$i = \frac{2}{2+1}*2= 1.33 A$

now power dissipated in 3 ohm resistance is given by

$P = i^2 * R$

$P = (1.33)^2* 3$

$P = 5.33 Watt$