Answer :
The potential difference across the first capacitor, which has a capacitance of 5.00 µF, is approximately 29.54 V when it is connected in series with another 8.00 µF capacitor across a 48.0 V source.
To calculate the potential difference across the first capacitor when a 5.00 µF and an 8.00 µF capacitor are connected in series across a 48.0 V potential difference, we first need to find the equivalent capacitance of the series combination. The equivalent capacitance (
C_{eq}) for capacitors in series is given by the reciprocal of the sum of the reciprocals of their individual capacitances:
\(1/C_{eq} = 1/C_1 + 1/C_2\)
\(1/C_{eq} = 1/5.00\mu F + 1/8.00\mu F\)
\(1/C_{eq} = 0.20\mu F^{-1} + 0.125\mu F^{-1}\)
\(1/C_{eq} = 0.325\mu F^{-1}\)
\(C_{eq} = 1/0.325\mu F^{-1}\)
\(C_{eq} = 3.077\mu F\)
Since the capacitors are in series, they share the same charge (
Q). The charge on a capacitor is given by \(Q = C \times V\). The total voltage across both capacitors is 48.0 V, so:
\(Q = C_{eq} \times V = 3.077\mu F \times 48.0 V\)
\(Q = 147.696\mu C\)
Now, we can find the potential difference across the first capacitor (V_1) by using the charge and its capacitance:
\(V_1 = Q / C_1 = 147.696\mu C / 5.00\mu F\)
\(V_1 = 29.5392 V\)
Therefore, the potential difference across the 5.00 µF capacitor is approximately 29.54 V.
