The capacitors have values C₁ = 2.0 μF and C₂ = 4.0 μF, C₂ = 5.0 μF C4 = 7.0 μF and the potential difference across the battery is 9.0 V. Assume that the capacitors are connected in series.
a) Find the equivalent capacitance of the circuit.
b) Solve for the potential difference across each capacitors.
Explanations & Calculations
a)
"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{1}{C_e}&=\\small \\frac{1}{2.0}+\\frac{1}{4.0}+\\frac{1}{5.0}+\\frac{1}{7.0}\\\\\n\\small C_e&=\\small 0.9\\, \\mu F\n\\end{aligned}"
b)
"\\qquad\\qquad\n\\begin{aligned}\n\\small Q&=\\small 0.9\\,\\mu F\\times 9.0\\,V\\\\\n&=\\small 8.1\\,\\mu C\\\\\n\\\\\n\\small V_1&=\\small \\frac{Q}{C_1}=\\frac{8.1\\,\\mu C}{2.0\\,\\mu F}=4.1\\,V\\\\\n\\small V_2 &=\\small \\frac{8.1}{4.0}= 2.1\\, V\\\\\n\\small V_3 &=\\small 1.6\\,V\\\\\n\\small V_4&=\\small 1.2\\,V\n\\end{aligned}"
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