Question

Draw a schematic diagram of a circuit consisting of 24v battery a 10ohm resistor a 5 ohm resistor a 1 oh resistor an ammeter and a plug key all connected in series find the ammeter and potential difference at ends of 1 ohm resistor

Answer 1

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