Answer :
When two capacitors of capacitance 6 μF and 4 μF are put in series across a 120 V battery, the potential difference across the 4 μF capacitor can be calculated using the formula
V2 = C2 / (C1 + C2) × V total, where V total is the total potential difference across the two capacitors, C1 is the capacitance of the first capacitor (6 μF), C2 is the capacitance of the second capacitor (4 μF), and V2 is the potential difference across the second capacitor (4 μF).So, substituting the given values, we get
V2 = 4 μF / (6 μF + 4 μF) × 120
V= 4 μF / 10 μF × 120
V= 48 V the potential difference across the 4 μF capacitor is 48 V.
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