Question

Two resister 5 ohm & 4 ohm insert in a circuit in Parallel , and two resister 8 ohm & 3 ohm insert in a series . count total resistance of this circuit??

Answer 1

Answer:

here is ur answer

Step-by-step explanation:

2.222ohm+2.182=4.404ohm

Answer 2

Answer

• Total resistance of this circuit will be = 13.22 ohm

Given :-

• Two resistance 5 ohm & 4 ohm connect in parallel .
• Other two resistance 8 ohm & 3 ohm connect with circuit in series .

Find :-

• Total resistance in circuit

Step - by - Step - Explanation

Assume that,

Resistance A & B are connected in parallel , and C & D connected in series.

Where,

• A = 5 ohm
• B = 4 ohm
• C = 8 ohm
• D = 3 ohm

Using Formula

Total Resistance when, resister connect in parallel ,

$\small{\boxed{\boxed{\tt{\red{\:\frac{1}{R_{t}}\:=\:\frac{1}{R_{1}}+\frac{1}{R_{2}}}}}}}$

Total Resistance when, resister connect in Series.

$\small{\boxed{\boxed{\tt{\red{\:R_{t'}\:=\:R_{1}+\:R_{2}}}}}}$

When, A & B connect in parallel,

➡1/Rt = 1/5 + 1/4

➡1/Rt = (5+4)/20

➡1/Rt = 9/20

Or,

➡Rt = 20/9

➡Rt = 2.22 ohm

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When, C & D connect in series,

➡Rt = 8 + 3

➡Rt' = 11 ohm

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But, Now Rt & Rt' also connected in series in this circuit.

So, Total resistance be,

➡ R = Rt + Rt'

➡R = 2.22 + 11

➡R = 13.22 ohm

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