Question

Two capacitors with C1 = 5.0 µF and C2 = 20.0 µF are initially uncharged and then connected in series to a battery. If neither capacitor can withstand a potential difference greater than 80 volts without breakdown, what is the maximum value of the battery emf that can be sustained by the circuit?

Answer 1

The maximum emf that the battery can sustain without causing either capacitor in series to break down is 100 volts. This is calculated by ensuring the voltage across the smaller capacitor does not exceed its maximum voltage rating and by using the equivalent series capacitance.

Step-by-step explanation:

When capacitors are connected in series, their voltages add up to equal the voltage of the battery. Each capacitor in a series must have the same charge (Q). However, because the capacitors have different capacitances, the voltage across each will vary. According to the formula V = Q/C, where V is the potential difference across the capacitor, Q is the charge, and C is the capacitance, a smaller capacitance will have a higher voltage across it for the same charge.

For two capacitors, C1 and C2, in series, we have:

• The total combined capacitance (Ctotal) is given by the formula 1/Ctotal = 1/C1 + 1/C2.
• The maximum voltage Vmax that the system can withstand is the smallest maximum voltage rating of the individual capacitors.
• To prevent breakdown of a capacitor, no capacitor should experience a voltage higher than its maximum voltage rating.

Considering the maximum potential difference across each capacitor is 80 volts, we can determine the maximum emf (Emax) the battery can have. First, calculate Ctotal:

1/Ctotal = 1/5.0 µF + 1/20.0 microF

Ctotal = 4.0 microF

Since during series connection the charge on each capacitor is the same and the sum of voltages across them should not exceed 80 V (the breakdown voltage of the smallest capacitor):

VC1 + VC2 = Emax

Because C1 is smaller it will have a higher voltage across it for the same charge, hence VC1 should be considered as 80 V to ensure neither capacitor breaks down:

Therefore, Emax = VC1 + (Q/C2),

and since, Q = C1 * VC1,

Emax = 80 V + ((5.0 microF * 80 V)/20.0 microF),

Emax = 80 V + 20 V = 100 V.

The maximum emf the battery can sustain without breaking down either capacitor is 100 volts.

Answer 2

100 volts.

Step-by-step explanation:

You haver to apply Kirchhoff’s Voltage Law for the given series circuit:

Let Vemf, VC1, VC2 represent the Voltage of the battery, the capacitor C1 and the capacitor C2 respectively.

Vemf = VC1 + VC2

Two capacitors in series can be expressed as one capacitor with a capacitance calculated as follow:

`(1)/(C)=(1)/(C1)+(1)/(C2)`

Solving for C and replacing the values of C1 and C2:

`C=(C1C2)/(C1+C2)=((5.0)(20.0))/(5.0+20.0)` = 4μF

The voltage divider formula for two capacitors in series is:

`VCn=Vs(C)/(Cn)`

Where Vs is the voltage of the battery, C is the equivalent capacitance of the capacitors in series and VCn is the voltage of one of the capacitors.

In this case:

`VC1=Vemf(C)/(C1)`

Solving for Vemf:

`Vemf=((VC1)(C1))/(C)`

The voltage (potential difference) of the circuit is proportional to the potencial difference of the capacitor. Therefore if the maximum value of the potential difference of any capacitor is 80 volts then the maximum value of Vemf is obtained using the maximum value of the voltage of the capacitor with the smaller capacitance (because that way you guarantee that the potential difference is not greater than 80 V for either capacitor. The reason is that the voltage of a capacitor is inversely proportional to its capacitance, so the capacitor with the greater value of voltage is C1)

`Vemf=((80)(5.0)(10^(-6)) )/((4.0)(10^(-6)))= 100 V`