Question

Two capacitances 0.5 µF and 0.75 µF are connected in parallel and the combination to a 110 V battery. Calculate the charge from the source and charge on each capacitor.

Answer 1

Hey There!!


Two capacitors, 0.5 $\mu F$ and 0.75 $\mu F$ are connected in parallel across a 110 V battery.


Our data is:

$C_1 = 0.5 \, \, \mu F \\ \\ C_2 = 0.75 \, \, \mu F \\ \\ V = 110 \, \, volts$



For two capacitors in parallel, the equivalent capacitance is given by:


$C_{eq} = C_1 + C_2 \\ \\ \\ \implies C_{eq} = 0.5 + 0.75 \\ \\ \\ \implies C_{eq} = 1.25 \, \, \mu F$


Now, Charge from Source can be calculated as follows:

$Q_{tot} = C_{eq} V \\ \\ \\ \implies Q_{tot} = (1.25 \, \, \mu F) \times (110 \, \, V) \\ \\ \\ \implies \boxed{Q_{tot}=137.5 \,\,\mu C}$

Here, tot simply represents the total amount of charge from source (i.e. battery).


Now, we know the total charge from source. 

In a parallel connection, the potential across each capacitor is same.  


Also, we have the formula $Q = CV$


Since we know that V will be constant, we see that Charge on each capacitor is directly proportional to its capacitance.


That is, $Q \propto C$

The total charge from source is going to be divided among the two capacitors. We can write:

$Q \propto C \\ \\ \\ \implies \frac{Q_1}{Q_{tot}} = \frac{C_1}{C_{eq}} \\ \\ \\ \implies Q_1 = Q_{tot} \times \frac{C_1}{C_{eq}} \\ \\ \\ \implies Q_1 = 1.375 \times \frac{0.5}{1.25} \\ \\ \\ \implies \boxed{Q_1 = 0.55 \, \, \mu C}$


Now, we also know the charge on the first capacitor. 

We also know that the charge from the source is divided among the two capacitors. So, we can write:

$Q_{tot} = Q_1 + Q_2 \\ \\ \\ \implies 1.375 = 0.55 + Q_2 \\ \\ \\ \implies Q_2 = 1.375 - 0.55 \\ \\ \\ \implies \boxed{Q_2 = 0.825 \, \, \mu C}$



Thus, we can summarize the answers as follows:

Charge from source = 1.375 $\mu C$
Charge on 0.5 $\mu F$ capacitor = 0.55 $\mu C$
Charge on 0.75 $\mu F$ capacitor = 0.825 $\mu C$



Hope it helps
Purva
Brainly Community

Answer 2

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