For the circuit in series:
E=100v
Resistor 1 = 3 kΩ
Resistor 2 = 4 kΩ
Resistor 3 = 1 kΩ, p= 200mW
Resistor 4 = R
find the following quantities:
a. The circuit currents
b. The total resistance of the circuit
c. The value of the unknown resistance, 6.
d. The voltage drops across all resistors in the circuit.
e. The power dissipated by all resistors.
3 answers
Again, one has to have a circuit diagram of how exactly the resistors are connected.
its in a series with 4 resistors and one cell battery
Given:
E = 130 Volts. 100v. is too low for the given circuit values.
R1 = 3k.
R2 = 4k.
R3 = 1k, P3 = 200mW.
R4 = ?
a. P3 = V3^2/R3 = 200.
V3^2/1 = 200,
V3 = 14.14 volts.
I3 = V3/R3 = 14.14/1=14.14 mA = I1 = I2 = I4.
b. V4 = E - I(R1+R2+R3) = 130 - 14.14mA*(8k) = 16.88 volts.
R4 = V4/I4 = 16.88/14.14 = 1.2k ohms.
Rt = 3 + 4 +1 + 1.2 = 9.2k.
c. R4 = 1.2k(see part b).
d. V1 = 14.14 * 3 = 42.42 Volts.
V2 = 14.14 * 4 = 56.56 Volts.
V3 = 14.14 * 1 = 14.14 Volts.
V4 = 14.14 * 1.2 = 16.97 Volts.
e. P1 = V1 * I = 42.42 * 14.14 = 600 mW.
P2 = 56.56 * 14.14 =
P3 = 14.14 * 14.14 =
P4 = 16.97 * 14.14 =
Note: All power in mW. All currents in mA.
E = 130 Volts. 100v. is too low for the given circuit values.
R1 = 3k.
R2 = 4k.
R3 = 1k, P3 = 200mW.
R4 = ?
a. P3 = V3^2/R3 = 200.
V3^2/1 = 200,
V3 = 14.14 volts.
I3 = V3/R3 = 14.14/1=14.14 mA = I1 = I2 = I4.
b. V4 = E - I(R1+R2+R3) = 130 - 14.14mA*(8k) = 16.88 volts.
R4 = V4/I4 = 16.88/14.14 = 1.2k ohms.
Rt = 3 + 4 +1 + 1.2 = 9.2k.
c. R4 = 1.2k(see part b).
d. V1 = 14.14 * 3 = 42.42 Volts.
V2 = 14.14 * 4 = 56.56 Volts.
V3 = 14.14 * 1 = 14.14 Volts.
V4 = 14.14 * 1.2 = 16.97 Volts.
e. P1 = V1 * I = 42.42 * 14.14 = 600 mW.
P2 = 56.56 * 14.14 =
P3 = 14.14 * 14.14 =
P4 = 16.97 * 14.14 =
Note: All power in mW. All currents in mA.
2 answers