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# As shown in the cireuit to the right, a potential difference of 625 V is applied...

As shown in the cireuit to the right, a potential difference of 625 V is applied across an air-filled parallel plate capacitor having capacitance Cr-2.36 x101l F and plate area 4-4.00x10m2. The dielectric constant of air is 1.00, Consider the switch in the circuit to be open as shown S 4 Find the charge on capacitor C (A)1.25 x10*C (B) 1.48 x10 c (C) 1.68x10 (D) 1.73x10-C (E) 1.93x10*c 625 v G, 6 Determine the between of the plates of capacitor Ci (A)1.20 x103m (B) 1.35 x103 m (C) 1.50 x103 m (D)1.65 x10m (E) 1.90 x10m Capacitor Cis uncharged when the switch is closed. Capacitor Cz is a parallel-plate capacitor having the same plate area and plate separation as capacitor Ci, and it is filled with polystyrene having a dielectric constant of 2.56. 1 6. How much charge passes through the switch after it is closed?? (A)3.17 x108 C (B)3.31x10 C (C)3.43x10*C (D)3.64x10* C (E)3.78x103 C What is the equivalent capacitance of the two capacitors in the circuit? (A)5.90x10-11 F (B)6.60x10-11 F (C)7.30x1011 F (D)8.40x10-11 F (B) 9.60x10-1 F Consider the circuit shown below. 200 0 Calculate the equivalent resistance of the circuit. (A)2.00 2 (B)2.20 2 (C) 2.40 ? (D)260 ? (E) 2.30 ? 3.00 0 1.00? 15.0 v Calculate the current through the 4.00 ? resistor. (A)2.00 A (B)2.50 A (C)3.00 A (D)3.50 A (E)4.00 A

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