Answer:
Sure, I can help you solve the problem for the power dissipated across the 3Ω resistor in the circuit.
The first step is to identify the current flowing through the 3Ω resistor. Since the 3Ω and 6Ω resistors are connected in parallel, the current through the 3Ω resistor will be the same as the current through the 6Ω resistor.
We can use the voltage divider rule to find the voltage across the parallel combination of the 3Ω and 6Ω resistors. The voltage divider rule states that the voltage across each resistor in a parallel combination is equal to the total voltage divided by the ratio of the total resistance to the resistance of that resistor.
In this circuit, the total voltage is 4 volts and the total resistance of the parallel combination is 3Ω + 6Ω = 9Ω. Therefore, the voltage across the parallel combination is 4 volts / 9Ω = 4/9 volts.
Since the voltage across the 3Ω and 6Ω resistors is the same, the current through the 3Ω resistor can be found using Ohm's law: I = V / R, where I is current, V is voltage, and R is resistance.
Therefore, the current through the 3Ω resistor is 4/9 volts / 3Ω = 4/27 amps.
Now that we know the current flowing through the 3Ω resistor, we can use Ohm's law again to find the power dissipated across the resistor. Power is equal to current squared times resistance (P = I^2 * R).
Therefore, the power dissipated across the 3Ω resistor is (4/27 amps)^2 * 3Ω = 30.45 watts.
So the answer is (1) 30.45 W.
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