A 20-Ω resistor and a 30-Ω resistor are wired in parallel and connected to a 9.0-V battery.
a. How much current is drawn from the battery?
b. How much power is dissipated in the circuit?
i = V/R
for 20-Ω
i = 9/20
for 30-Ω
i = 9/30
so
total current from battery = 9/20 + 9/30 = 27/60 + 18/60 = 45/60 = 3/4 amp
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or use parallel resistor formula
1/R = 1/R1 + 1/R2
R = R1 R2 /(R1+R2) = 600 / 50 = 12 Ohms
i = V/R = 8/12 = 3/4 amp again
Power = volts * amps = 9 * 3/4 = 27/4 watts
a. Well, the current drawn from the battery can be calculated using Ohm's Law. But let's not be so "Ohm" about it, shall we? Imagine the 20-Ω resistor is a water slide, and the 30-Ω resistor is a roller coaster. These two resistance rides are wired in parallel, which means it's the same as having two amusement parks side by side. So, when you have options, you usually run for the water slide, right? More people choose the path of least resistance! In this case, the 20-Ω resistor provides an easier path for current, so most of the current will flow through it. Therefore, the current drawn from the battery will be more influenced by the 20-Ω resistor, and only a small fraction will go through the 30-Ω resistor.
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.
To find the current drawn from the battery and the power dissipated in the circuit, you can use Ohm's law and the power formula. Here's how you can calculate it:
a. To find the current (I) drawn from the battery, you can use Ohm's law which states that current is equal to voltage divided by resistance. In this case, we have two resistors connected in parallel, so the total resistance (R_total) is given by the formula:
1/R_total = 1/R1 + 1/R2
Let's calculate it:
1/R_total = 1/20 + 1/30
1/R_total = (3 + 2)/60
1/R_total = 5/60
R_total = 60/5
R_total = 12 Ω
Now that we know the total resistance, we can use Ohm's law to calculate the current:
I = V/R_total
I = 9.0 V / 12 Ω
I ≈ 0.75 A
Therefore, the current drawn from the battery is approximately 0.75 Amperes.
b. To find the power dissipated in the circuit, you can use the power formula:
P = I^2 * R
Let's calculate it for each resistor separately:
For the 20 Ω resistor:
P1 = I^2 * R1
P1 = (0.75 A)^2 * 20 Ω
P1 ≈ 11.25 W
For the 30 Ω resistor:
P2 = I^2 * R2
P2 = (0.75 A)^2 * 30 Ω
P2 ≈ 16.87 W
Therefore, the total power dissipated in the circuit is the sum of both resistors' power:
Total Power = P1 + P2
Total Power ≈ 11.25 W + 16.87 W
Total Power ≈ 28.12 W
So, the power dissipated in the circuit is approximately 28.12 Watts.