A 10.0 Ω resistor is hooked up in series with an 8.0 Ω resistor followed by a 27.0 Ω resistor. The circuit is powered by a 9.0 V battery
-Calculate the equivalent resistance.
-Calculate the voltage drop across each resistor in the circuit.
R = 10 + 8 + 27
i = 9/R
for each Vn = i * Rn
the sum of Vn better equal 9
Well, I must say, these resistors are quite resistant to telling jokes! But I'll do my best to calculate everything for you without any resistance.
To calculate the equivalent resistance (Req), we simply add up the resistances in the series circuit. So, Req = 10.0 Ω + 8.0 Ω + 27.0 Ω = 45.0 Ω.
Now, let's calculate the voltage drop across each resistor. Since they are in a series circuit, the total voltage of 9.0 V will be divided among the resistors.
Using Ohm's law (V = I * R), we can calculate the current (I) flowing through the circuit first. Since it's a series circuit, the current remains the same across all resistors. So, I = V / Req. Plugging in the values, we get I = 9.0 V / 45.0 Ω = 0.2 A.
Now, we can calculate the voltage drop across each resistor using V = I * R.
For the 10.0 Ω resistor, V1 = I * R1 = 0.2 A * 10.0 Ω = 2.0 V.
For the 8.0 Ω resistor, V2 = I * R2 = 0.2 A * 8.0 Ω = 1.6 V.
And for the 27.0 Ω resistor, V3 = I * R3 = 0.2 A * 27.0 Ω = 5.4 V.
So there you have it! The equivalent resistance is 45.0 Ω, and the voltage drop across each resistor is 2.0 V, 1.6 V, and 5.4 V respectively.
To calculate the equivalent resistance of resistors in series, you simply add their individual resistances.
In this case, the equivalent resistance is:
10.0 Ω + 8.0 Ω + 27.0 Ω = 45.0 Ω
Now let's calculate the voltage drop across each resistor.
The total voltage in a series circuit is equal to the sum of the voltage drops across each resistor. In this case, the total voltage is 9.0 V.
To calculate the voltage drop across each resistor, you can use Ohm's Law (V = I * R), where V is the voltage, I is the current, and R is the resistance. Since the resistors are in series, the current will be the same through each resistor.
First, calculate the total current using Ohm's Law:
I = V / R_total
I = 9.0 V / 45.0 Ω
I = 0.2 A
Now, calculate the voltage drop across each resistor:
Voltage drop across 10.0 Ω resistor: V = I * R
V = 0.2 A * 10.0 Ω
V = 2.0 V
Voltage drop across 8.0 Ω resistor:
V = I * R
V = 0.2 A * 8.0 Ω
V = 1.6 V
Voltage drop across 27.0 Ω resistor:
V = I * R
V = 0.2 A * 27.0 Ω
V = 5.4 V
Therefore, the voltage drop across each resistor in the circuit is as follows:
- 10.0 Ω resistor: 2.0 V
- 8.0 Ω resistor: 1.6 V
- 27.0 Ω resistor: 5.4 V
To calculate the equivalent resistance in a series circuit, you simply add up the resistances of all the components in the circuit. In this case, the equivalent resistance (R_eq) is given by:
R_eq = R1 + R2 + R3
where R1, R2, and R3 are the resistances of the three resistors respectively.
In this circuit, R1 = 10.0 Ω, R2 = 8.0 Ω, and R3 = 27.0 Ω. Therefore, the equivalent resistance is:
R_eq = 10.0 Ω + 8.0 Ω + 27.0 Ω
R_eq = 45.0 Ω
Now, to calculate the voltage drop across each resistor, we can use Ohm's Law, which states that the voltage drop across a resistor (V) is equal to the product of the current (I) flowing through the resistor and the resistance (R) of the resistor.
V = I * R
Since the circuit is powered by a 9.0 V battery, the total voltage across the circuit is 9.0 V. Therefore, the current flowing through each resistor (I) is the same.
Now, let's calculate the current (I) using Ohm's Law:
I = V / R_eq
I = 9.0 V / 45.0 Ω
I = 0.2 A
Therefore, the current flowing through each resistor in the circuit is 0.2 A.
Now, we can calculate the voltage drop across each resistor using Ohm's Law and the current (I) we just calculated.
For the 10.0 Ω resistor:
V1 = I * R1
V1 = 0.2 A * 10.0 Ω
V1 = 2.0 V
For the 8.0 Ω resistor:
V2 = I * R2
V2 = 0.2 A * 8.0 Ω
V2 = 1.6 V
For the 27.0 Ω resistor:
V3 = I * R3
V3 = 0.2 A * 27.0 Ω
V3 = 5.4 V
Therefore, the voltage drop across the 10.0 Ω resistor is 2.0 V, across the 8.0 Ω resistor is 1.6 V, and across the 27.0 Ω resistor is 5.4 V.