Grade 12Electrostatics

three capacitors c1=20mf c2=40mf c3=50mf are connected in series .if no capacitor can sustain more than 50v ,then the maximum potential difference between the two terminals X and Y is?

shweta bhardwaj

13 Years agoGrade 12

2 Answers

Arun Kumar

12 Years ago

In series charges are same.
$Q/C_1+Q/C_2+Q/C_3=Q/C_{res} \\=>V_i=Q/C_i \\=>max(V_i)=50$
We get Q from here then we will get V by RHS.
$V=Q/C_{res}$
Thanks & Regards
Arun Kumar
IIT Delhi
Askiitians Faculty

Kushagra Madhukar

6 Years ago

Dear student,

Please find the solution to your problem below.

Let, the charge on each capacitor be q.

Now, Potential drop across a capacitor, V = Q/C

Hence, the potential drop across the given capacitors will be-

For C₁, V₁ = q/20

For C₂, V₂ = q/40

For C₃, V₃ = q/50

Now, V₁ is the highest among all the individual voltages.

Its max value can reach upto 50 volts

Hence, q/20 = 50

or, q = 1000 mC

Hence, V₁ = 50 V

V₂ = 1000/40 = 25 V

V₃ = 1000/50 = 20 V

Hence potential difference between x and y terminal, V = V₁ + V₂ + V₃

= 50 + 25 + 20

= 95 V

Thanks and regards,

Kushagra

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three capacitors c1=20mf c2=40mf c3=50mf are connected in series .if no capacitor can sustain more than 50v ,then the maximum potential difference between the two terminals X and Y is?

shweta bhardwaj

2 Answers

Arun Kumar

Kushagra Madhukar

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