Question

Question 27 (6 points) 27. For the circuit shown find: (a) the equivalent resistance of the circuit. (b) The current though 240 resistor on the right branch of the circuit. www 4.0 6.0 24 V WW LW 24 02 WW 8.00 24 w C Format V в Іо ...

Answer 1

The circuit is redrawn below.

Ri=42 D 3 C R=662 V=24V LH R=240 R₂=803 Rue 242 a F D W

(a) From the figure, R3 and R4 are in parallel combination. Let their equivalent resistance be R(A). It is given by,

$R_{A}=\frac{R_{3}R_{4}}{R_{3}+R_{4}}=\frac{8\times 24}{8+24}=6\Omega$

Now R(A) and R2 are in series combnation. Let their equialent resistance be R(B). It is given by,

RB = R2+ R = 6+ 6 = 1212

Now R(B) and R5 are in parallel combination. Let their equivalent resistance be R(C). It is given by,

$R_{C}=\frac{R_{5}R_{B}}{R_{5}+R_{B}}=\frac{24\times 12}{24+12}=8\Omega$

Now R(C) and R1 are in series combination. Their equivalent resistance is the equivalent resistance of the circuit. It is given by,

Req R1 + Rc= 4+ 8 = 1212

So the equivalent resistance of the circuit is 12$\Omega$.

(b) The total current in the circuit is,

$I=\frac{V}{R_{eq}}=\frac{24}{12}=2A$

Now this current split into two parts at the node B in the figure. By current division formula, the current through R5 is,

$I_{5}=\frac{IR_{B}}{R_{5}+R_{B}}=\frac{2\times 12}{24+12}=0.67A$

So the current through the 24$\Omega$ resistor on the right branch of the circuit is 0.67A.