Question

In the diagram below

Answer 1

Answer:

ANSWER

As in the given circuit,

R

1

and R

2

are in parallel,

So, R

1

∥R

2

=

R

1

+R

2

R

1

⋅R

2

⇒Req1=

10+40

10×40

=

50

400

=8Ω

And, R

3

,R

4

and R

5

are in parallel,

So,

R

3

∥R

4

∥R

5

=

R

3

1

+

R

4

1

+

R

5

1

=

R

eq2

1

R

eq2

1

=

30

1

+

20

1

+

60

1

=

60

2+3+1

=

10

1

∴R

eq2

=10Ω

Now, R

eq1

and R

eq2

are in series,

∴R

eq1

+R

eq2

=R

eq

⇒R

eq

=8Ω+10Ω=18Ω

we know,

I=

R

eq

V

=

18×10

1.8

=

180

18

=0.1A

Therefore, ammeter reading of the given circuit is 0.1A

solution

Step-by-step explanation:

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