Question

(a) Find the equivalent resistance of the network
shown below
[Ans: 2.18 2]
(b) What is the current in 89 resister if the po-
tential difference of 12V is applied to the net-
work?
(Ans: 0.56 A]
w
522
602
892
352
12v.
13,9 A 60V potential difference is applied to the circuit​

Answer 1

Answer:

a)  2.2 Ω (Approx)

b)$\\\tt \approx 0.5A$

Explanation:

a)  Since 5Ω and 8 Ω are in parallel,

$\\\tt \dfrac{1}{R_{eq}} = \dfrac{1}{5 \Omega} +\dfrac{1}{8 \Omega} \\R_{eq} = 3.07 \Omega$

Since 6 Ω is in series with the above equivalent

Net R here = 9.07Ω

Now this thing is parallel with 3Ω

net R using same methods as in  5Ω and 8 Ω equals 2.2 Ω (Approx)

b) Voltage in lower 3Ω will be same as in the line with 6Ω and equivalent of  5Ω and 8 Ω

Current in the the line with 6Ω and equivalent of  5Ω and 8 Ω =  $\\\tt I =\dfrac{12}{R_{net}} = \dfrac{12 V}{9.07 \Omega} = 1.3 2A$

Now the current in 5Ω and 8 Ω  combination will be same as in 6 Ω,

PD across the combination = $\\\tt I \times R_{net} = 1.3 \times 3.07=4.05 V$

Since they're both in parallel both have same PD across them.

Now,

Current though 8  Ω = $\\\tt I = \dfrac{V}{R} =\dfrac{4.05 V}{8 \Omega} \approx 0.5A$

Answer 2

Solution :-

1]

We know that

$\sf \dfrac{1}{n} = \dfrac{1}{R1}+\dfrac{1}{R2}$

1/n = 1/5 + 1/8

1/n = 8 + 5/40

1/n = 13/40

1/n =  3.07 ohm

Now

When 6 added = 9.07 ohm

Now

With 3 ohm will occur same 6 ohm

Hence it will become 2.2 ohm

2] Finding the current

V = IR

12 = I(9.07)

12/9.07 = I

1.32 = I

1.32 × 3.07 = 4.0524

Now

V = IR

4.0524 = I(8)

4.0524/8 = I

0.56 = I