Answer to Question #110282 in Electric Circuits for James

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

The electric circuit in the condition of the problem looks like shown in the figure.

Solution (a): The first parallel connection of resistances $R_1$ and $R_2$ has a common resistance

(1) $R_{12}=\frac{R_1\cdot R_2}{R_1+R_2}=\frac{15\cdot 10}{15+10}\Omega=6\Omega$. For the second parallel connection, we have a similar formula

(2) $R_{3x}=\frac{R_3\cdot R_x}{R_3+R_x}$ . The impedance of the circuit is the sum

(3) $R=R_{12}+R_{3x}$ since $R_{12}$ and $R_{3x}$ are connected in series. The total current from power supply is (4) $I=\frac{V}{R}$, and the total power dissipated

(5) $W=V\cdot I$ . Substituting (3) and (4) to (5) and find R we get

$W=\frac{V^2}{R}$ or $R=\frac{V^2}{W}=\frac{250^2\cdot V^2}{2.5\cdot 10^3W}=25 \Omega$

Substituting this value and (1) (2) to (3) we get

$25\Omega=6\Omega+\frac{R_3\cdot R_x}{R_3+R_x}$ Deciding on $R_x$ we get

$R_x=\frac{R_3\cdot 19}{R_3-19}=\frac{38\cdot 19}{38-19}==38\Omega$

Solution (b): The total current from power supply is (4) with $R=25\Omega$ we have

(6) $I=\frac{250V}{25\Omega}=10A$ This current flows through both series resistance $R_{12}$ and $R_{3x}$ , so these parts of the circuit can be considered separately. In the second parallel combination, both resistances $R_3$ and $R_x$ included in the parallel are the same; the same current flows through them $I_3=I_x=I/2=5A$ . In the first part of the circuit, the voltage at both resistances is equal to

$V_1=I\cdot R_{12}=10A\cdot 6\Omega=60V$ This voltage is applied to both the first and second resistance. therefore

$I_1=\frac{V_1}{R_1}=\frac{60V}{15\Omega}=4A$

$I_2=\frac{V_1}{R_2}=\frac{60V}{10\Omega}=6A$

Answer: (a) The value of the resistor Rx, such that the total power dissipated in the circuit is 2.5kW is 38 Ohms,

Answer: (b)The current flowing in each of the four resistors is $I_1=4A, I_2=6A, I_3=5A, I_x=5A$