To find the magnitude of the current flowing through the first device in a parallel circuit, we can use Ohm's Law and the property of parallel circuits.
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Calculate the total resistance (R_total) of the parallel circuit: The formula for the total resistance \( R_{total} \) of devices in parallel is given by: \[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \] where \( R_1 \) is the resistance of the first device (12 ohms) and \( R_2 \) is the resistance of the second device (4 ohms).
\[ \frac{1}{R_{total}} = \frac{1}{12} + \frac{1}{4} \] To add these fractions, we find a common denominator, which is 12: \[ \frac{1}{R_{total}} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3} \] Therefore, \[ R_{total} = 3 \text{ ohms} \]
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Calculate the total current (I_total) from the generator: Using Ohm's Law, where \( V = IR \), \[ I_{total} = \frac{V}{R_{total}} = \frac{40 , V}{3 , \Omega} = \frac{40}{3} \approx 13.33 , A \]
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Calculate the current through the first device (I_1): Since the voltage across devices in parallel is the same, it is equal to the voltage of the generator (40 V). Using Ohm's Law again for the first device: \[ I_1 = \frac{V}{R_1} = \frac{40 , V}{12 , \Omega} \approx 3.33 , A \]
So, the magnitude of the current flowing through the first device is approximately 3.33 A.
Thus, the correct answer is:
A) 3.33 A