Question

. In an electrical circuit two resistors of 2Ω and 4Ω respectively are connected in

series to a 6V battery. The heat dissipated by the 4Ω resistor in 5 s is

(a) 5J (b) 10J (c) 20J (d) 30J

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Answer 1

Answer:

20 joules

Explanation:

Energy dissipated = Pt or VI×t or IsquareRT 

Consider Ohm's law V=IR.

In this case the resistors are connected in series and hence, the total resistance is 4+2=6 ohms. In a series circuit, the current remains the same throughout the circuit and so we use the Isquare Rt Joules formula.

The total current in the circuit is calculated as I=V/R=6/6=1ampere.

Therefore, the power dissipated P across the 4 ohm resistor for 5 s

= 1  

1 ×4×5=20Joules

Hence, The heat dissipated by the 4 resistor in 5 s will be 20 J

hope it helps u..............