Four resistors are connected to a battery with a terminal voltage of 12.0 V, (R1 = 40.0 , R2 = 80.0 , R3 = 140.0 and R4 = 15.0 )
(a) Determine the equivalent resistance for the circuit.
1
(b) Determine the current in the battery.
2 V
(c) Determine the current in the 40.0 resistor.
3 A
(d) Determine the power dissipated by the 80.0 resistor.
4 W
(e) Determine the power dissipated by the 15.0 resistor.
5 W
(Hint: Remember that P= (ÄV)2/R = IÄV.)
_ i don't understand how to do this at all??? HELP
No worries! I'm here to help you step-by-step. Let's break down each question one by one.
(a) To determine the equivalent resistance for the circuit, we can use the formula for resistors connected in parallel:
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4
Substituting the given values:
1/Req = 1/40 + 1/80 + 1/140 + 1/15
To simplify this further, we can find the least common multiple (LCM) of the denominators of the fractions, which is 840:
1/Req = (21/840) + (10/840) + (6/840) + (56/840)
1/Req = 93/840
To find Req, we take the reciprocal of both sides:
Req = 840/93 ≈ 9.03 Ω
Therefore, the equivalent resistance for the circuit is approximately 9.03 Ω.
(b) To determine the current in the battery, we can use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R):
I = V/R
Substituting the given values:
I = 12.0/9.03
I ≈ 1.33 A
Therefore, the current in the battery is approximately 1.33 A.
(c) To determine the current in the 40.0 Ω resistor, we can use Ohm's Law again:
I = V/R
Substituting the given values (V = 12.0 V, R = 40.0 Ω):
I = 12.0/40.0
I = 0.30 A
Therefore, the current in the 40.0 Ω resistor is 0.30 A.
(d) To determine the power dissipated by the 80.0 Ω resistor, we can use the formula for power:
P = (ΔV)^2/R
Substituting the given values (ΔV = 12.0 V, R = 80.0 Ω):
P = (12.0)^2/80.0
P = 1.8 W
Therefore, the power dissipated by the 80.0 Ω resistor is 1.8 W.
(e) To determine the power dissipated by the 15.0 Ω resistor, we can use the same power formula:
P = (ΔV)^2/R
Substituting the given values (ΔV = 12.0 V, R = 15.0 Ω):
P = (12.0)^2/15.0
P = 9.6 W
Therefore, the power dissipated by the 15.0 Ω resistor is 9.6 W.
I hope this helps! Let me know if you have any further questions.
To solve this problem, we will use Ohm's Law and the formulas for calculating electrical power.
(a) To find the equivalent resistance of the circuit, we need to consider how the resistors are connected.
In this case, the four resistors are connected in series, which means that the total resistance is equal to the sum of the individual resistances:
Rs = R1 + R2 + R3 + R4
Substituting the given values:
Rs = 40.0 + 80.0 + 140.0 + 15.0 = 275.0 Ω
So, the equivalent resistance for the circuit is 275.0 Ω.
(b) To determine the current in the battery, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R):
I = V / R
Substituting the given terminal voltage (V = 12.0 V) and the equivalent resistance (R = 275.0 Ω):
I = 12.0 V / 275.0 Ω = 0.0436 A
Therefore, the current in the battery is 0.0436 A, or 43.6 mA (milliamperes).
(c) To determine the current in the 40.0 Ω resistor, we can use the same formula as in part (b):
I = V / R
Substituting the given terminal voltage (V = 12.0 V) and the individual resistance (R = 40.0 Ω):
I = 12.0 V / 40.0 Ω = 0.3 A
So, the current in the 40.0 Ω resistor is 0.3 A.
(d) To determine the power dissipated by the 80.0 Ω resistor, we can use the formula for electrical power:
P = I^2 * R
Substituting the current (I = 0.0436 A) and the resistance (R = 80.0 Ω):
P = (0.0436 A)^2 * 80.0 Ω = 0.151 W
Therefore, the power dissipated by the 80.0 Ω resistor is 0.151 W, or 151 mW (milliwatts).
(e) To determine the power dissipated by the 15.0 Ω resistor, we can use the same formula as in part (d):
P = I^2 * R
Substituting the current (I = 0.0436 A) and the resistance (R = 15.0 Ω):
P = (0.0436 A)^2 * 15.0 Ω = 0.028 W
So, the power dissipated by the 15.0 Ω resistor is 0.028 W, or 28 mW.
I hope this explanation helps you understand how to solve this circuit problem. Let me know if you have any further questions!
Assuming the load is not too great, and that the resistors are in parallel, then
1/R = 1/40 + 1/80 + 1`/140 + 1/15
R = 9
I = V/R = 12/9 = 1.3 A
I(40) = 12/40 = .3 A
P(80) = 144/80 = 1.8 W
P(15) = 144/15 = 9.6 W