In a series circuit, there are 4 resistor with one battery cell. (there's no diagram for this question)
The battery cell = 100v
r1 = 3kΩ
r2 = 4kΩ
r3 = 1kΩ, P=100mW
r4 = ?
find the following quantities:
1) the circuit currents
2)the total resistance of the circuits
3)the value of the unknown resistance of r4
4) the voltage drops across all resistors
5) the power dissipated by all resistors
On r3, get current by P=I^2/R
Then if you circuit current, so you know Rtotal (100/current)
then you know R4 (Rtotal-r1-r2-r3)
then you know V4 (IR4)
total power= I^2 * Rtotal or you can add each IV up for all four resistors.
can you show me how to do it
Given:
E = 100 Volts.
R1 = 3k.
R2 = 4k.
R3 = 1k, P3 = 100mW.
R4 = ?
1. I^2 * R3 = 100mW.
I^2 * 1 = 100,
I = 10mA. = I1 = I2 = I3 = I4.
2. R = E/I = 100/10mA = 10k Ohms.
3. R4 = R - (R1+R2+R3) = 10k - 8k = 2k Ohms.
4. V1 = I*R1 = 10 *3 = 30 Volts.
V2 = I*R2 = 10 * 4 = 40 Volts.
V3 = I*R3 =
V4 = I*R4 =
5. P1 = V1*I = 30 * 10 = 300 mW.
P2 = V2*I = 40 * 10 = 400 mW.
P3 = V3*I =
P4 = V4*I =
thanks henry2,. you are a lifesaver
Glad I could help.
To solve this problem, we need to apply the principles of Ohm's Law and Kirchhoff's Laws to analyze the series circuit. Let's go through each question step by step:
1) To find the circuit current, we can use Ohm's Law, which states that I = V/R, where I is the current, V is the voltage, and R is the resistance. Since the resistors are connected in series, the current flowing through each resistor will be the same. Therefore, the circuit current can be calculated using any of the resistors. Let's use r1 for this calculation:
I = V / R
I = 100V / 3kΩ
I ≈ 0.033A or 33mA (rounded to 3 decimal places)
So, the circuit current is approximately 33mA.
2) The total resistance of a series circuit is simply the sum of all the individual resistances. In this case, we have four resistors: r1, r2, r3, and r4. So, the total resistance can be found by adding their values:
Total Resistance (R_total) = r1 + r2 + r3 + r4
R_total = 3kΩ + 4kΩ + 1kΩ + r4
R_total = 8kΩ + r4
Unfortunately, we do not have the value for r4, so we cannot calculate the total resistance without it.
3) To find the value of the unknown resistance r4, we need additional information or equations. The problem statement doesn't provide any specific details or equations related to r4. Therefore, at this point, we cannot determine the value of r4.
4) The voltage drop across each resistor in a series circuit is proportional to its resistance. To find the voltage drops across all resistors, we need to multiply the circuit current by each resistor's resistance:
Voltage Drop across r1 = I * r1
Voltage Drop across r1 = 0.033A * 3kΩ
Voltage Drop across r1 = 99V
Voltage Drop across r2 = I * r2
Voltage Drop across r2 = 0.033A * 4kΩ
Voltage Drop across r2 = 132V
Voltage Drop across r3 = I * r3
Voltage Drop across r3 = 0.033A * 1kΩ
Voltage Drop across r3 = 33V
We don't have the value of r4, so we cannot calculate the voltage drop across it.
5) The power dissipated by each resistor can be calculated using the formula P = (I^2) * R, where P is power, I is current, and R is resistance. Let's calculate the power dissipated by each resistor:
Power dissipated by r1 = (I^2) * r1
Power dissipated by r1 = (0.033A)^2 * 3kΩ
Power dissipated by r1 ≈ 0.0333W or 33.3mW (rounded to 4 decimal places)
Power dissipated by r2 = (I^2) * r2
Power dissipated by r2 = (0.033A)^2 * 4kΩ
Power dissipated by r2 ≈ 0.044W or 44mW (rounded to 3 decimal places)
Power dissipated by r3 = (I^2) * r3
Power dissipated by r3 = (0.033A)^2 * 1kΩ
Power dissipated by r3 ≈ 0.011W or 11mW (rounded to 3 decimal places)
We don't have the value of r4, so we cannot calculate the power dissipated by it.
In summary, we were able to find the circuit current, voltage drops and power dissipation values for the given resistors. However, we cannot determine the total resistance or the value of the unknown resistor r4 without more information.