Question

10
6. In an electrical circuit two resistors of 2 1 and 4 1 respectively are
connected in series to a 6 V battery. The heat dissipated by the 4 resistor
in 5 s will be *​

Answer 1

Explanation:

Energy dissipated = PtorVI×tor =I

2

Rt Joules

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 I

2

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

2

×4×5=20Joules

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