Question

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

Answer 1

Answer:

h = vit

so we have to find i

I=v/r

l=6/4

l= 3/2

apply to formula

h=vit

h=6×3/2×5

h=45joules

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Answer 2

Answer:

The heat dissipated by the 4-ohm resistor is 45J.

Explanation:

The heat dissipated by the resistor is given as,

$H=I^2Rt$           (1)

Where,

H=heat dissipated by the resistor

I=current through the resistor

R=resistance of the resistor

t=time during which the current flows through the resistor

From the question we have,

R₁=2Ω

R₂=4Ω

The voltage of the battery(V)=6volt

t=5 seconds

Since both the resistors 2Ω and 4Ω are connected in parallel the voltage across them will remain the same i.e.6 volt.

So, the current through the 4Ω resistor will be,

$I=\frac{V}{R}=\frac{6}{4} =\frac{3}{2} A$      (2)

By substituting the required values in equation (1) we get;

$H=(\frac{3}{2}) ^2\times 4\times 5$

$H=45 Jsec$

Hence, the heat dissipated by the 4-ohm resistor is 45J.