Answer to Question #168475 in Electric Circuits for Chaihcai

Question

Given:

1st Resistor = 5 ohm

2nd Resistor = 12 ohm

3rd resistor = 6 ohm

4th resistor = 6 ohm

5th resistor = 30 ohm.

1. Suppose the four resistors are connected in parallel. What is the resistance of the combined resistors?

2. Suppose the four resistors are connected in parallel. Find (a) equivalent resistance (b) current flowing through each resistor (c) voltage across each resistor and (d) power dissipated by each resistor.

3. Find the equivalent resistance of the combination of resistance shown.

Answer 1

Explanations & Calculations

• The equivalent resistance of a set of series-connected resistors is given by

$\qquad\qquad \begin{aligned} \small R_e&= \small R_1+R_2+R_3+.. \end{aligned}$

• And that of a parallel-connected set

$\qquad\qquad \begin{aligned} \small \frac{1}{R_e}&= \small \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+... \end{aligned}$

• When there are only two resistors in parallel, their equivalence is simply given by

$\qquad\qquad \begin{aligned} \small \frac{1}{R_e}&= \small \frac{1}{R_1}+\frac{1}{R_2}\\ &= \small \frac{R_1+R_2}{R_1\times R_2}\\ \\ \small R_e&= \small \frac{R_1\times R_2}{R_1+R_2} \end{aligned}$

• As the information for the first two parts is insufficient (and it is the same you had requested through the task), I guess you need the answer to the third part of the question.
• For those types of questions, first, you need to identify a convenient end of the circuit to start with (most probably a branch with only two connections: no more).
• Try to find the equivalence of sets of resistors as much as possible & proceed.

• To understand, refer to the sketch attached where the circuit is step by step broken down to the simplest form.

• Both the 6 Ohm resistors are in a parallel arrangement, hence the equivalent is

$\qquad\qquad \begin{aligned} \small &= \small \frac{6\times6}{6+6}=\bold{3\Omega} \end{aligned}$

• Then the 12 Ohm & the 3 Ohm are in a series combination whose equivalent is

$\qquad\qquad \begin{aligned} \small &= \small 12+3=\bold{15\Omega} \end{aligned}$

• Now, 15 Ohm & the 30 Ohm resistors are in a parallel arrangement whose equivalent is

$\qquad\qquad \begin{aligned} \small &= \small \frac{15\times30}{15+30}=\bold{10 \Omega} \end{aligned}$

• Then that 10 Ohm & the 5 Ohm are in a series & their equivalent is

$\qquad\qquad \begin{aligned} \small &= \small 5+10=\bold{15\Omega} \end{aligned}$

• Finally, the equivalence of the entire circuit is $\small \bold{15\Omega}$

• At each step, the equivalent resistance found is encircled with a dashed line