Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
Share
Notifications

View all notifications

Two Capacitors of Capacitance 20⋅0 Pf and 50⋅0 Pf Are Connected in Series with a 6⋅00 V Battery. Find (A) the Potential Difference Across Each Capacitor and (B) the Energy Stored in Each Capacitor. - Physics

Login
Create free account


      Forgot password?

Question

Two capacitors of capacitance 20⋅0 pF and 50⋅0 pF are connected in series with a 6⋅00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Solution

Given : 

`C_1 = 20.0  "pF"`

`C_2 = 50.0  "pF"`

When the capacitors are connected in series, their equivalent capacitance is given by `C_(eq) = (C_1C_2)/(C_1+C_2)`

∴ Equivalent capacitance,`C_(eq) = ((50 xx 10^-12) xx (20 xx 10^-12))/((50 xx 10^-12)+(20 xx 10^-12))` = `1.428 xx 10^-11  "F"`

(a) The charge on both capacitors is equal as they are connected in series. It is given by 

`q = C_(eq) xx V`

⇒ `q = (1.428 xx 10^-11) xx 6.0  "C"`

Now , 

`V_1 = q/C_1 = ((1.428 xx 10^-11) xx 6.0  "C")/((20 xx 10^-12))`

⇒ `V_1 = 4.29  "V"`

and

`V_2 = (6.00 - 4.29) V = 1.71   "V"`

(b) The energies in the capacitors are given by

`E_1 = q^2/(2C_1)`

= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 20 xx 10^-12)`

= `184  "pJ"`

and

`E_2 = q^2/(2C_1)`

= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 50 xx 10^-12)`

= `73.5  "pJ"`

  Is there an error in this question or solution?

APPEARS IN

Solution Two Capacitors of Capacitance 20⋅0 Pf and 50⋅0 Pf Are Connected in Series with a 6⋅00 V Battery. Find (A) the Potential Difference Across Each Capacitor and (B) the Energy Stored in Each Capacitor. Concept: Electric Potential Difference.
S
View in app×