Question

17.Calculate (a) the value of the resistor Rx, such that the total power dissipated in the circuit is 2.5KW, and (b) the current flowing in each of the four resistors . R1=15ohms, R2=10ohms, R3=38ohms and R4=Rx.
Note that R1 and R2 are connected parallel and R3 and Rx are also separately connected in parallel. The resistors are connected to 250 V power supply. (Draw the circuit diagram)

Answer 1

a) The current in the circuit is: $I = P/V = 2500/250 = 10$ A. The voltage across the resistors 1 and 2 by tne Ohm's law is: $U_{12} = I\cdot \dfrac{R_1R_2}{R_1+R_2} = 10\cdot \dfrac{15\cdot10}{15+10} = 60$ V.

The voltage across the resistors 3 and 4 is:

$U_{34} = U - U_{12} = 90$ V.

The resistance of the resistor $R_4$ can be found from the Ohm's law:

$U_{34} = I\cdot \dfrac{R_3R_4}{R_3+R_4} = 10\cdot \dfrac{38\cdot R_4}{R_4+38} =90$

$R_4 = 11.8$ Ohm

b) The currents in the resistors:

$I_1 = U_{12}/R_1 = 60/15 = 4$ A;

$I_2 = U_{12}/R_2 = 60/10 = 6$ A;

$I_3 = U_{34}/R_3 = 90/38 \approx2.37$ A;

$I_4 = U_{34}/R_4 = 90/11.8 \approx 7.63$ A.

Answer. a) 11.8 Ohm b) 4A, 6A, 2.37A, 7.63A.