Consider the combination of capacitors in the figure below. (Let C, = 28.0 mu F
ID: 2233508 • Letter: C
Question
Consider the combination of capacitors in the figure below. (Let C, = 28.0 mu F and C2 = 2.00 mu F.) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance. In diagram 1 find the equivalent capacitance of the three capacitors in parallel and redraw the diagram as a single battery and single capacitor in a loop. Compute the charge on the single equivalent capacitor. Returning to diagram 1, compute the charge on each individual capacitor. Does the sum agree with the value found in part (c)? What is the charge on the 28.0- mu F capacitor and on the 8- mu F capacitor? Compute the voltage drop across the 28.0- mu F capacitor. Compute the voltage drop across the 2.00- mu F capacitor.Explanation / Answer
a) equivalent of the series is 6.22 microF b)now the three capacitors are in parallel so equivalent C = 6.22 + C2 + 4 = 12.22 micro F c)charge Q = VC = 36 * 12.22 = 440 micro Coulomb d) charge on : 4 micro F cpacitor = V * 4 = 144 micro Coulomb 4 micro F cpacitor = V * 2 = 72 micro Coulomb 6.22 micro F cpacitor = V * 6.22 = 223.92 micro Coulomb yes the sum agree with the value found in part c e) as this two capacitors are in series value of charge is same as there equivalent Q = =223.92 micro Coulomb f)voltage drop across 28 micro F cpacitor = Q/C = 223.92/28 = 7.99 Volt g)voltage drop across 2 micro F cpacitor = Q/C = 72/2 = 36 volt = voltage of battery