Physics, asked by abhisheksahni174, 3 days ago

Q.38. Ammeter reading in the given circuit will be 3 Ω 3 Ω 3 Ω 3 Ω 1 Ω 2 Ω + HE A 12 Ω (α) 2 Α (6) 1A (c) 1.5A (d) 0.5 A.​

Answers

Answered by jmargaretdaniel
1

Answer:

(a)As all the resistances are in parallel the voltage across each of them will be same =12V

I  

1

=  

R  

1

 

V

=  

5

12

=2.4A

I  

2

=  

R  

2

 

V

=  

10

12

=1.2A

I  

3

=  

R  

3

 

V

=  

30

12

=0.4A

(b) Total current= I  

1

+I  

2

+I  

3

=2.4+1.2+0.4=4A

(c) Total circuit resistance =  

Total current

V

=  

4

12

=3Ω

Explanation:

Answered by p231
1

Answer:

Labelling each of the resistance as shown in the figure below.

Resistance R₁ and R₂ are in series

⇒ R$1= R₁ + R₂

⇒ R$1 = 3 + 3 = 6 Ω

Now R1 and R3 are in parallel.

1 Rp 1 1 + R$1 R3

1 Rp 1 1 + 6 3 介

介 1 Rp 1+2 6 3 36

6 3 ⇒Rp = 2

⇒ Rp = 20

Now R4, Rp and R5 are in series.

⇒ Req = R4 + R5 + Rp

Explanation:

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