Question
Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit.
✓
Answer
$\because\frac{\text{C}_{1}}{\text{C}_{2}} = \frac{\text{C}_{3}}{\text{C}_{4}}$
This is the condition of balance so there will be no current across PR (50 μF capacitor)
Now $?_1$ and $?_2$ are in series,
$\text{C}_{12} = \frac{\text{C}_{1}\text{C}_{2}}{\text{C}_{1} + \text{C}_{2}} = \frac{10\times20}{10 + 20 } = \frac{200}{30} =\frac{20}{3}\mu\text{F}$
$\because$ $?_3$ and $?_4$ are in series,
$\text{C}_{34} = \frac{\text{C}_{3}\text{C}_{4}}{\text{C}_{3} + \text{C}_{4}} =\frac{5\times10}{5 + 10 } =\frac{50}{15} = \frac{10}{3}\mu\text{F}$
Equivalent capacitance between A and B is
$\text{C}_{AB} = \text{C}_{12} + \text{C}_{34} =\frac{20}{3} + \frac{10}{3} = 10 \mu\text{F}$
Charge drawn from battery (q) = CV
$ = 10 \times 10 \mu\text{C}$
$ = 10\mu\text{C} \text{ or }10^{-4}\text{ C}.$
This is the condition of balance so there will be no current across PR (50 μF capacitor)
Now $?_1$ and $?_2$ are in series,
$\text{C}_{12} = \frac{\text{C}_{1}\text{C}_{2}}{\text{C}_{1} + \text{C}_{2}} = \frac{10\times20}{10 + 20 } = \frac{200}{30} =\frac{20}{3}\mu\text{F}$
$\because$ $?_3$ and $?_4$ are in series,
$\text{C}_{34} = \frac{\text{C}_{3}\text{C}_{4}}{\text{C}_{3} + \text{C}_{4}} =\frac{5\times10}{5 + 10 } =\frac{50}{15} = \frac{10}{3}\mu\text{F}$
Equivalent capacitance between A and B is
$\text{C}_{AB} = \text{C}_{12} + \text{C}_{34} =\frac{20}{3} + \frac{10}{3} = 10 \mu\text{F}$
Charge drawn from battery (q) = CV
$ = 10 \times 10 \mu\text{C}$
$ = 10\mu\text{C} \text{ or }10^{-4}\text{ C}.$
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