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?

100 volts.

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:

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

= 4μF

The voltage divider formula for two capacitors in series is:

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:

Solving for Vemf:

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)

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j