Question

draw a schematic diagram of circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor and a 15 ohm resistor , and a plug key, all are connected in series.Calculate the electric current passing through the above above circuit when the key is closed.

Answer 1

Answer:

Explanation:

As per the data provided in the sum three resistors of 5Ω,10Ω and 15Ω are connected in series so the net resistance = 5Ω +10Ω+15Ω = 30Ω.

    Also given that five cells of 2 V are connected in series,

So, the total voltage of the battery is 10V,according to Ohm's law  

V=IR    and  I=V/R    

On substituting resultant voltage(V) as 10 V and resultant resistance as 30Ω.

       I=10 V/ 30Ω. = 3.33 ampere