Question

state the steady value of ready of ammeter of R=1 ohm and C=1/π F​

Answer 1

Answer:

Ans is 48 v

Explanation:

In steady state current through capacitor wire is zero. So ammeter reading A

1

is zero and current flow through 200Ω,900Ω,A

2

.

Potential across the capacitor is V

c

=V

1

=

C

q

=

100×10

−6

4×10

−3

=40V

Now this is potential across the total resistance =900+ resistance of A

2

=900+100=1000Ω

Thus the reading of A

2

is I=

1000

V

1

=40/1000=(1/25)A

As 1000Ω and 200Ω are in series so current through is equal to I.

Thus, V

2

=V

200

=200I=200/25=8V

Emf of cell is E=V

1

+V

2

=40+8=48V

Answer 2

Answer:

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