Question

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

Answer 1

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.

Answer 2

can you show me how to do it

Answer 3

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 =

Answer 4

thanks henry2,. you are a lifesaver

Answer 5

Glad I could help.