Each of the four circuits in the drawing consists of a single resistor whose resistance is either R or 2R, and a single battery whose voltage is either V or 2V. The unit of voltage in each circuit is V = 10 V and the unit of resistance is R = 8 Ω. Determine (a) the power supplied to each resistor and (b) the current delivered to each resistor.
You say each circuit has one resistor.
(a) The power in that resistor is
P = V^2/R
(b) The current in the resistor is
I = V/R
V is either 10 or 20 volts, depending upon the circuit.
R is either 8 or 16 Ù, depending upon the circuit.
Why did the circuit go to therapy? Because it had some serious resistance issues!
(a) Let's calculate the power supplied to each resistor. Remember that power (P) is given by the equation P = I^2 * R, where I is the current and R is the resistance.
In Circuit 1:
Since the resistance is R and the voltage is V, we have R = 8Ω and V = 10V.
The current (I) in Circuit 1 can be found using Ohm's law, which states that V = I * R. So, I = V / R = 10V / 8Ω = 1.25A.
Now we can calculate the power supplied to the resistor in Circuit 1: P1 = I^2 * R = (1.25A)^2 * 8Ω = 12.5W.
In Circuit 2 and Circuit 3:
The resistance is 2R = 2 * 8Ω = 16Ω.
The voltage is 2V = 2 * 10V = 20V.
Using Ohm's law again, we find the current in both circuits: I = V / R = 20V / 16Ω = 1.25A.
The power supplied to the resistors in Circuit 2 and Circuit 3 is the same as in Circuit 1: P2 = P3 = (1.25A)^2 * 16Ω = 25W.
In Circuit 4:
The resistance is R = 8Ω.
The voltage is 2V = 2 * 10V = 20V.
Using Ohm's law once more, we find the current in Circuit 4: I = V / R = 20V / 8Ω = 2.5A.
The power supplied to the resistor in Circuit 4 is: P4 = (2.5A)^2 * 8Ω = 50W.
So, the power supplied to each resistor is:
Circuit 1: 12.5W
Circuit 2: 25W
Circuit 3: 25W
Circuit 4: 50W
(b) Now, let's determine the current delivered to each resistor.
In Circuit 1, we already found that the current (I) is 1.25A.
In Circuits 2 and 3, the current is also 1.25A.
In Circuit 4, the current (I) is 2.5A.
So, the current delivered to each resistor is:
Circuit 1: 1.25A
Circuit 2: 1.25A
Circuit 3: 1.25A
Circuit 4: 2.5A
I hope this electrifying explanation sparked a smile!
To determine the power supplied to each resistor, we can use the formula:
Power = (Voltage)^2 / Resistance
(a) Power supplied to each resistor in each circuit:
In circuit 1, the resistor has resistance R and the battery has voltage V.
Power = (V)^2 / R = (10V)^2 / (8Ω) = 12.5W
In circuit 2, the resistor has resistance 2R and the battery has voltage V.
Power = (V)^2 / (2R) = (10V)^2 / (2 * 8Ω) = 6.25W
In circuit 3, the resistor has resistance R and the battery has voltage 2V.
Power = (2V)^2 / R = (20V)^2 / (8Ω) = 50W
In circuit 4, the resistor has resistance 2R and the battery has voltage 2V.
Power = (2V)^2 / (2R) = (20V)^2 / (2 * 8Ω) = 25W
(b) Current delivered to each resistor in each circuit:
To find the current delivered to each resistor, we can use Ohm's Law:
Current = Voltage / Resistance
In circuit 1, the resistor has resistance R and the battery has voltage V.
Current = V / R = 10V / 8Ω = 1.25A
In circuit 2, the resistor has resistance 2R and the battery has voltage V.
Current = V / (2R) = 10V / (2 * 8Ω) = 0.625A
In circuit 3, the resistor has resistance R and the battery has voltage 2V.
Current = (2V) / R = (2 * 10V) / 8Ω = 2.5A
In circuit 4, the resistor has resistance 2R and the battery has voltage 2V.
Current = (2V) / (2R) = (2 * 10V) / (2 * 8Ω) = 1.25A
So, the power supplied to each resistor and the current delivered to each resistor in each circuit are as follows:
Circuit 1:
- Power = 12.5W
- Current = 1.25A
Circuit 2:
- Power = 6.25W
- Current = 0.625A
Circuit 3:
- Power = 50W
- Current = 2.5A
Circuit 4:
- Power = 25W
- Current = 1.25A
To determine the power supplied to each resistor and the current delivered to each resistor, we can use the formulas:
1. Power (P) = Voltage (V) * Current (I)
2. Ohm's Law: Voltage (V) = Current (I) * Resistance (R)
Let's analyze each circuit separately:
Circuit 1:
- The resistance is R = 8 Ω.
- The voltage is V = 10 V.
To find the current (I), we use Ohm's Law: I = V / R = 10 V / 8 Ω = 1.25 A.
To find the power (P), we use P = V * I = 10 V * 1.25 A = 12.5 W.
So, in Circuit 1, the power supplied to the resistor is 12.5 W, and the current delivered to the resistor is 1.25 A.
Circuit 2:
- The resistance is 2R = 2 * 8 Ω = 16 Ω.
- The voltage is V = 10 V.
Using Ohm's Law, we find the current (I): I = V / R = 10 V / 16 Ω = 0.625 A.
The power (P) is given by P = V * I = 10 V * 0.625 A = 6.25 W.
In Circuit 2, the power supplied to the resistor is 6.25 W, and the current delivered to the resistor is 0.625 A.
Circuit 3:
- The resistance is R = 8 Ω.
- The voltage is 2V = 2 * 10 V = 20 V.
Using Ohm's Law, we find the current (I): I = V / R = 20 V / 8 Ω = 2.5 A.
The power (P) is given by P = V * I = 20 V * 2.5 A = 50 W.
In Circuit 3, the power supplied to the resistor is 50 W, and the current delivered to the resistor is 2.5 A.
Circuit 4:
- The resistance is 2R = 2 * 8 Ω = 16 Ω.
- The voltage is 2V = 2 * 10 V = 20 V.
Using Ohm's Law, we find the current (I): I = V / R = 20 V / 16 Ω = 1.25 A.
The power (P) is given by P = V * I = 20 V * 1.25 A = 25 W.
In Circuit 4, the power supplied to the resistor is 25 W, and the current delivered to the resistor is 1.25 A.
Thus, the power and current values for each resistor in the circuits are as follows:
Circuit 1: Power = 12.5 W, Current = 1.25 A
Circuit 2: Power = 6.25 W, Current = 0.625 A
Circuit 3: Power = 50 W, Current = 2.5 A
Circuit 4: Power = 25 W, Current = 1.25 A