Question

800
ure

36. Two resistors 400 ohm And 800 ohm
connected in series with a battery 6V.
An ammeter of 10 ohm resistance is used
to measure current in circuit. What be reading of of ammeter
(a)4.96 mA (b) 5.83 mA
(c)7.25 mA (d)3.86m A
​

Answer 1

Answer:

total resistance=400+800=1200ohm

ammeter reading I=V/R

I=6/1200

I=0.005

please mark it as brainliest

Answer 2

Solution :-

Given

▪️Value of resistors = 400 ohm and 800 ohm

▪️Type of circuit = series

▪️Value of Resistance if Ammeter = 10 ohm .

▪️Potential Difference across circuit = 6V

Now as it is a series circuit then :-

R equivalent = 400 + 800

= 1200 ohm

Now as there is also Resistance in Ammeter , it will be added in series.

Then Final R equivalent

= 1200 + 10

= 1210 ohm

Now as V = IR

=> 6 = I × 1210

=> 6 ÷ 1210 = I

=> I = 0.004958

=> I = 0.0046

=> I = 4.6 mA