Question

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

Answer 1

Answer:

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

Explanation:

Energy dissipated = Pt or,

VI×t or,

=I^2Rt 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^2Rt 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.

Answer 2

Answer:

20 joule...

Explanation:

...

Since the 2 ohm and 4 ohm are in series so equivalent resistance is 6 ohm.

Using formula V=I×R

6=I × 6

I=6/6=1 Ampere.

Now heat dessipated through 4 ohm = H=I²RT

H= 1×1×4×5 ( Since resistance is 4 ohm,current is 1 Ampere and time is 5 second)

H=20 Joule...