Questions
ln the circuit shown in figure the potential difference across the 4.5 F capacitor is
detailed solution
Correct option is A
Let us calculate the total capacitance of the circuit. The capacitors of capacitances 3 μF and 6 μF arc in parallel, Their capacitance C' is given by C′=3μF+6μF=9μFThe capacitor of capacitance C' is in series with the capacitor of capacitance 4.5 μF. Therefore, the capacitance of the system is given by 1C=19+14⋅5=13or C=3μFThe charge flowing through the circuit=3×12=36μC∴Potential difference a cross 4.5 μF capacitor=qC=364⋅5=8 voltTalk to our academic expert!
Similar Questions
When two capacitors one of 3 F and the other of 6 F arc connected in series and the combination is charged to a potential difference of 120 volt, the potential difference across 3 F capacitor is
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