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
