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In an electrical circuit, two resistors of 2 ohms and 4 ohms, respectively, are connected in parallel to a 6 V battery. What is the heat dissipated by the 4-ohm resistor in 5 seconds?

Answer :

Final answer:

The 4-ohm resistor connected in parallel with a 2-ohm resistor to a 6V battery dissipates 45 Joules of heat in 5 seconds.

Explanation:

Calculation of Heat Dissipated

To determine the heat dissipated by the 4-ohm resistor when connected in parallel with a 2-ohm resistor to a 6V battery, we can use Ohm's law and the formula for power dissipation.

First, we find the total resistance (RT) in the circuit when the resistors are in parallel:
RT = 1 / (1/2Ω + 1/4Ω) = 4/3Ω

Next, we calculate the total current (IT) using Ohm's law (V = IR):
IT = V / RT = 6V / (4/3Ω) = 4.5A

Now, since the voltage is the same across resistors in parallel, we can calculate the current through the 4-ohm resistor (I4Ω):
I4Ω = V / R4Ω = 6V / 4Ω = 1.5A

The power dissipated by the 4-ohm resistor is given by the formula P = I2R, where P is power, I is current, and R is resistance:
P4Ω = I4Ω2 × R4Ω = (1.5A)2 × 4Ω = 9W

The heat dissipated (Q) over time (t) in seconds can be found by multiplying power by time:
Q = P4Ω × t = 9W × 5s = 45 Joules

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