Three capacities of capacitance C_1=7.00pf,c_2=6.00ph and c_3=3.00pf are connect

Question

Three capacities of capacitance C_1=7.00pf,c_2=6.00ph and c_3=3.00pf are connected in series. A potential difference of delta V = 11.0 V is maintained by a battery find the equivalent capacitance of the series of capacitors and the change on each capacitor Determine the effect on the equivalent capacitance of reducing the second capacitance to 0.1 times its previous value a charge of +q on the capacitor plate connected of the assembly to the positive battery terminal required a charges of -q on the opposite plate Q=Q_1=Q_2=Q_3 The sum of the potential differences across all three capacitors equals the potential of the battery delta v=delta V_1+delta v_1+ delta V_2+delta V_3, and taking eavh delta N to be Q/c LENDS TO THE RELATION 1/C_equals=1/c_1+1/c_2+1/c_3 so equals=1/1/c_1+1/c_2+1/c_3=1/(1/7.00 pf+1/6.00 pf+1/3.00pf) and the charge q is then gicen by q=v_equals delta V=

Explanation / Answer

1/Ceq =1/C1+1/C2+1/C3=1/7+1/6+1/3 =>Ceq =1.555pF

Q=Q1=Q2=Q3 =Ceq*V =1.555*11=17.11pC

2) C2 =0.1*6 =0.6pF

=>1/Ceq =1/7+1/0.6+1/3=> Ceq =0.467pF

Ceq decreases by 1.555-0.467 =1.088pF

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