Question

resistance of the wire.
8. Find the equivalent resistance of the following system and total current passi
through it.
(Ans. 422, 3 Amp)
#how?​

Answer 1

Answer:

Given:

• Given circuit diagram
• V = 12V

To find:

• Equivalent resistance
• Total current

$\mathfrak{\underline{Solution:}}$

In the given diagram, I have marked the points for clarification.

Clearly , the resistors between A and B are parallel combination. For solving resistance between A and B

⚕Formula used:

$\boxed{\dfrac{1}{R_{p}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

☘Putting the values,

$R_1 = R_2 = R_3 = 6$Ω

➫$\dfrac{1}{R_{AB}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}$

➫$\dfrac{1}{R_{AB}} = \cancel{\dfrac{3}{6_{{}_{{}_{2}}}}}$

➫$\dfrac{1}{R_{AB}} = \dfrac{1}{2}$

➫$R_{AB} = 2$Ω

Now the resistance from A to D is in series combination. Solving for resistance between A and D

⚕Formula used:

$\boxed{R_s = R_1 + R_2}$

☘Putting all the values:

$R_1 = R_{AB} = 2 Ω , \ R_2 =R_4 = 4 Ω$

➫$R_{AD} = 2 + 4$

➫$R_{AD} = 6$Ω

Now again , the resistance between C and E is in parllel combination. So , using the above said formula ,

☘Putting the values:

$R_1 = R_{AD} = 6 Ω, \ R_2 = R_5 =12 Ω$

➫$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{AD}} + \dfrac{1}{12}$

➫$\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{12}$

➫$\dfrac{1}{R_{eq}} = \dfrac{2 + 1}{12}$

➫$\dfrac{1}{R_{eq}} = \cancel {\dfrac{3}{12_{{}_{{}_{4}}}}}$

➫$\dfrac{1}{R_{eq}} = \dfrac{1}{4}$

➫$R_{eq} = 4$Ω

Hence the equivalent resistance is 4 Ω

✦Now finding the current :

⚕Formula to be used:

$\boxed{I = \dfrac{V}{R}}$

☘Putting vales in this Ohm's Law:

➫$I = \cancel{\dfrac{12^{{}^{{}^{3}}}}{4}}$

➫$I = 3 \ A$