Problen 10: Three capacitors are connected as shown in the figure. C,-32 C,-13.6

Question

Problen 10: Three capacitors are connected as shown in the figure. C,-32 C,-13.6pEd,-£9. The voltage on the battery is 12 V. C1 C3 C2 Otheexpertta.com Part (a) Express the equivalent capacitance of the two capacitors Cy and Cy in terms of the variables given in the problem statement. Expression 12 Select from the variables below to write your expression. Note that all variables may not be required. , , AV, 0, C, CI,C12-C2-Cg, d, g, h, m, P, t Part (b) Using the above result, express the total capacitance in terms of C1 and C Expression Select from the variables below to write your expression. Note that all variables may not be required. a, B, AV, 6, C, C C12. C2. C3. d. g, h, m, P, t Part (c) Calculate the numerical value of the total capacitance in F Numeric :A numeric value is expected and not an expression. Part (d) Express the charge Q stored in the circuit in terms of capacitance C and the potential difference V across the battery. Expression Select from the variables below to write your expression. Note that all variables may not be required. , , AV, 0, C,c,,c,2,C3, c3, d, g, h, m, P, t Part (e) Calculate the numerical value of i Numerie :A numeric value is expected and not an expression ress the energy stored in a capacitor in terms of capacitance C and the potential difference V. Expression: Select from the variables below to write your expression. Note that all variables may not be required. , , AV, 0. a, C, CI,C12-C2. C3, d, g, h, j, k, m, P, S. t, V Part (g) Calculate the numerical value of U in J. Numeric : A numeric value is expected and not an expression. U=

Explanation / Answer

here C1 = 3.2 uF = 3.2 x 10-6 F C2 = 13.6 uF = 13.6 x 10-6 F C3 = 5.9 uF = 5.9 x 10-6 F

and delta V = 12 volt

(a) here C1 and C2 are in parallel so their equilent capacitance

C12 = C1 + C2

= 16.8 x 10-6 F Ans

(b) here now C12 and C3 are in series so their equilent capacitance

C = C12 || C3

C = ( C12 x C3 ) / ( C12 + C3 ) Ans

(c) so the numerical value of total capacitance

C = ( 16.8 x 10-6 x 5.9 x 10-6 ) / [ (16.8 x 10-6) + (5.9 x 10-6) ]

= (99.12 x 10-12) / (22.7 x 10-6)

C = 4.366 x 10-6 F Ans

(d) as we know that the charge stored in the capacitor

C = Q / delta V

then Q = C x delta V Ans

(e) so the numerical value of charge

Q = 4.366 x 10-6 x 12

Q = 52.398 x 10-6 C = 52.398 uC Ans

(f) as we know that the energy stored in the capactor

E = ( Q x delta V ) / 2

here Q = C x delta V

then E = ( C x delta V2 ) / 2 Ans

(g) so the numerical value of energy

E = ( 4.366 x 10-6 x 12 x 12 ) / 2

E = 314.352 x 10-6 J = 314.352 uJ Ans

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