Question

13. For the circuit shown below, calculate:

(a) total resistance in the arm CE,

(b)total current drawn from the battery, and

(c) current in each arm, i.e., AB and CE of the circuit.

Answer 1

................................................................................................................✡✡✡
A) TOTAL RESISTANCE IN THE ARM CE = CD + DE

TOTAL RESISTANCE IN THE ARM CE = 2 OHM + 4 OHM

TOTAL RESISTANCE IN THE ARM CE = 6 OHM
................................................................................................................✡✡✡
B) ➡ CD AND DE FORMS SERIES CONNECTION

SO, CE = 6 OHM

➡ CE AND AB FORMS A PARALLEL CONNECTION

SO, 1 / R = 1 / R1 + 1 / R2

➡ 1 / R = 1 / 6 + 1 / 6

➡ 1 / R = 2 / 6

➡ R = 6 / 2

➡ R = 3 OHM

➡ TOTAL RESISTANCE OF THE CIRCUIT = 3 OHM

➡ VOLTAGE OF THE BATTERY
= 3V

➡ TOTAL CURRENT DRAWN FROM THE BATTERY = VOLT / RESISTANCE

➡ TOTAL CURRENT DRAWN FROM THE BATTERY = 3 / 3

➡ TOTAL CURRENT DRAWN FROM THE BATTERY = 1 AMPERE
................................................................................................................✡✡✡
C) ➡ CURRENT IN AB = VOLT / RESISTANCE

➡ CURRENT IN AB = 3 / 6

➡ CURRENT IN AB = 1 / 2 (0.5) AMPERES

➡ CURRENT IN CE = VOLT / RESISTANCE

➡ CURRENT IN CE = 3 / 6

➡ CURRENT IN CE = 1 / 2 (0.5) AMPERES
................................................................................................................✡✡✡
➡ HOPE IT HELPS

➡ GOOD NIGHT MY FRIEND

Answer 2

Answer:

The resistance in the arm CE is $6\Omega$ .

Current passing through the circuit is 1 ampere .

The current passing through the circuit arm AB and CE is 0.5 ampere .

Step-by-step explanation:

First Part

As shown in the figure given in the question .

Let

$R_{1}=2\Omega$

$R_{2}=4\Omega$

Thus

$Resistance\ in\ the\ arm\ CE = R_{1}+R_{2}$

$Resistance\ in\ the\ arm\ CE =2\Omega+4\Omega$        

$Resistance\ in\ the\ arm\ CE =6\Omega$        

Therefore the resistance in the arm CE is $6\Omega$        

Second Part

As shown in the figure

Let

$R_{3}=6\Omega$

Thus

$Total\ resistamce\ in\ the\ battery = \frac{1}{R_{3}}+\frac{1}{R_{CE\ arm}}$

Put all the values in the above

$\frac{1}{Total\ resistamce\ in\ the\ battery}= \frac{1}{6}+\frac{1}{6}$

$\frac{1}{Total\ resistamce\ in\ the\ battery}= \frac{2}{6}$

$\frac{1}{Total\ resistamce\ in\ the\ battery}= \frac{1}{3}$

$Total\ resistance\ in\ the\ battery = 3\Omega$

Now by using the ohms law

V = IR

Where V is the voltage , I is the current and R is the resistance .

As shown in the figure

V = 3v

Thus

$3 = I\times 3$

$I = \frac{3}{3}$

I = 1 ampere

Thus current passing through the circuit is 1 ampere .

Third part

[tex]Current\ in\ arm\ AB\ of\ the\ circuit =\frac{3}{6}[\tex]

[tex]Current\ in\ arm\ AB\ of\ the\ circuit =\frac{1}{2}[\tex]

                                                                   = 0.5 ampere

[tex]Current\ in\ arm\ CE\ of\ the\ circuit =\frac{3}{6}[\tex]

[tex]Current\ in\ arm\ CE\ of\ the\ circuit =\frac{1}{2}[\tex]

                                                                   = 0.5 ampere

Therefore the current passing through the circuit arm AB and CE is 0.5 ampere .