To solve this problem, we need to apply Ohm's Law and the power formula for resistors.
a. To find the current drawn from the battery, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R).
In this case, the total resistance of the parallel combination of the two resistors is given by the formula:
1/RTotal = 1/R1 + 1/R2,
where R1 and R2 are the resistances of the individual resistors.
Using this formula, the total resistance (RTotal) is calculated as follows:
1/RTotal = 1/20 Ω + 1/30 Ω.
To simplify the calculations, we can find the common denominator and then take the reciprocal of both sides:
1/RTotal = (3/60 + 2/60) / 60 = 5/60 / 60 = 5/360,
RTotal = 360/5 = 72 Ω.
Now, we can use Ohm's Law to find the current drawn from the battery:
I = V / RTotal,
I = 9.0 V / 72 Ω,
I ≈ 0.125 A.
Therefore, the current drawn from the battery is approximately 0.125 A.
b. To find the power dissipated in the circuit, we use the power formula for resistors, which states that the power (P) dissipated in a resistor is equal to the current (I) flowing through it squared, multiplied by its resistance (R).
For each resistor, we can calculate the power as follows:
P1 = I^2 * R1,
P2 = I^2 * R2.
Substituting the known values into the formula, we get:
P1 = (0.125 A)^2 * 20 Ω,
P2 = (0.125 A)^2 * 30 Ω.
Calculating these values:
P1 = 0.125^2 * 20 = 0.125 * 0.125 * 20 = 0.25 W,
P2 = 0.125^2 * 30 = 0.125 * 0.125 * 30 = 0.375 W.
Finally, we can add these powers to find the total power dissipated in the circuit:
P = P1 + P2,
P = 0.25 W + 0.375 W,
P ≈ 0.625 W.
Therefore, the power dissipated in the circuit is approximately 0.625 W.