Question
For the circuit shown in the adjoining Fig. $13.37$, the charge on $4 \mu \mathrm{F}$ capacitor is(A) $30 \mu \mathrm{C}$(B) $40 \mu \mathrm{C}$(C) $24 \mu \mathrm{C}$(D) $54 \mu \mathrm{C}$
Step 1
The equivalent capacitance $C_{eq}$ of capacitors connected in parallel is given by the sum of the individual capacitances. Therefore, we have \[C_{eq} = C_1 + C_2 = 1 \mu F + 5 \mu F = 6 \mu F.\] Show more…
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For the circuit shown in figure the charge on $4 \mu \mathrm{F}$ capacitor is (A) $20 \mu \mathrm{c}$ (B) $24 \mu \mathrm{c}$ (C) $30 \mu \mathrm{c}$ (D) $54 \mu \mathrm{c}$
For the circuit shown in figure, charge on the $5 \mu \mathrm{F}$ capacitor is: (a) $100 \mu \mathrm{C}$ (b) $250 \mu \mathrm{C}$ (c) $300 \mu \mathrm{C}$ (d) $500 \mu \mathrm{C}$
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