Consider the combination of capacitors in the figure below. (Let C1 = 26.0 mu F and C2 = 0.500 mu F.) (a) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance. ______________ mu F (b) 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. _____________ - mu F (c) Compute the charge on the single equivalent capacitor. _____________ - mu C (d) Returning to diagram 1, compute the charge on each individual capacitor. 4.00- mu F capacitor ____________ mu C 0.500- mu F capacitor ____________ mu C far right capacitor __________ mu C Does the sum agree with the value found in part (b)? Yes No (e) What is the charge on the 26.0- mu F capacitor and on the 8- mu F capacitor? ________ mu C (f) Compute the voltage drop across the 26.0- mu F capacitor. ___________V (g) Compute the voltage drop across the 0.500- mu F capacitor. ___________V
Explanation / Answer
from the diagram C1 = 26F 8.0F are in series the resultant capactence 1/Ceq = 1/C1 + 1/8F = 1/26F + 1/8F Ceq = 6.117F from the diagram Ceq , C2 and 4F are connected in parallel the resultant capacitence CR = Ceq+ C2 + 4F =6.11F+0.5F+4F =10.61F charge on the single equlent capaciotr Q= CR V = 10.61F(36v) =381.96C d) charge on 4C capacitor Q1 = 4C( 36v) =144C charge on the 0.5C capacitor Q2 = 0.5C ( 36v) =18c as far rigth capacioor equalence capacitence Q3 = Ceq v = 6.11F ( 36v) =219.9C Q(total) = Q1+Q2+ Q3 yes sum agree the charge on 26F and 8F are connected in series so these two capacitors have same charge which the equlent charge 219.9C voltage across the 26F capaciotr V1 = Q3/C1 = 219.9C/26F =8.46v voltgae across 8C capacitor V2 = Q3/8F =219.9/8Fv =27.4v from the diagram C1 = 26F 8.0F are in series the resultant capactence 1/Ceq = 1/C1 + 1/8F = 1/26F + 1/8F Ceq = 6.117F from the diagram Ceq , C2 and 4F are connected in parallel the resultant capacitence CR = Ceq+ C2 + 4F =6.11F+0.5F+4F =10.61F charge on the single equlent capaciotr Q= CR V = 10.61F(36v) =381.96C d) charge on 4C capacitor Q1 = 4C( 36v) =144C charge on the 0.5C capacitor Q2 = 0.5C ( 36v) =18c as far rigth capacioor equalence capacitence Q3 = Ceq v = 6.11F ( 36v) =219.9C Q(total) = Q1+Q2+ Q3 yes sum agree the charge on 26F and 8F are connected in series so these two capacitors have same charge which the equlent charge 219.9C voltage across the 26F capaciotr V1 = Q3/C1 = 219.9C/26F =8.46v voltgae across 8C capacitor V2 = Q3/8F =219.9/8Fv =27.4v = 219.9C/26F =8.46v voltgae across 8C capacitor V2 = Q3/8F =219.9/8Fv =27.4v