  Question

Combination of two identical capacitors, a resistor R and a DC voltage source of voltage 6V is used in an experiment on C-R circuit. If is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 s. For series combination the time needed for reducing the voltage of the fully charged series combination by half is

A
20 s  B
10 s  C
5 s  D
2.5 s  Solution

The correct option is D 2.5 sFor CR circuit, during charging the voltage across capacitor at instant t is $$V=V_0(1-e^{-t/CR})$$According to question, $$\frac{V_0}{2}=V_0(1-e^{-t/CR})$$or $$e^{t/CR}=2 \Rightarrow t=CRln2$$For parallel grouping , equivalent capacitance , $$C_P=C+C=2C$$now time for half voltage , $$t_P=C_PR ln2=2CR ln2$$For series grouping , equivalent capacitance , $$C_S=\frac{CC}{C+C}=C/2$$now time for half voltage , $$t_S=C_SR ln2=(C/2)R ln2$$$$\therefore \frac{t_S}{t_P}=\frac{1}{4} \Rightarrow t_S=\frac{1}{4}\times 10=2.5 s$$   as $$t_P=10 s$$ PhysicsNCERTStandard XII

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