Question

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

Answer 1

Answer:2o Joules

Explanation:

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  ×4×5=20Joules

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

Answer 2

Answer:

Ans. (c) 20J

Explanation: Total resistance of combination = 2ohm+4ohm= 6ohm

Current through the circuit can be calculated as follows:

$i = \frac{v}{r}$

which is =6v/6ohm=1A

Heat dissipation by 4ohm can be calculated as follows:

H=I²Rt = (1.4)² x 4ohm x 5s = 20J