1. A parallel circuit consists of 18uF capacitor (C1), a 29uF capacitor (C2) and a 225V battery.What is the charge of each capacitor on the circuit?2. An 88uF capacitor is charged to a voltage of 530 volts. How many coulombs of electriccharge are stored in this capacitor?3. Three 36µF capacitors are connected in series with each other. What will be theequivalence capacitance?4. For the series circuit of 4 capacitors with capacitances C1=1.5F, C2=2.2F, C3=6.9F, C4=8.6Fand a battery with 1000V find the charges and voltages for each capacitor.5. Find the total resistance in number 4 in 1 day in scientific notation.6. Calculate the value of capacitance needed to store 3.5μC of charge at 256V.
Answer (1)
Let's solve these capacitor circuit problems step by step.1. Parallel Circuit: C1 = 18µF, C2 = 29µF, V = 225VIn a parallel circuit, the voltage across each capacitor is the same as the battery voltage. The charge (Q) on a capacitor is given by the formula: (Q = CV), where C is the capacitance and V is the voltage.Charge on C1 (Q1):(Q1 = C1 \times V = 18 , \mu F \times 225 , V)(Q1 = 4050 , \mu C)(Q1 = 4.050 \times 10^{-3} , C)Charge on C2 (Q2):(Q2 = C2 \times V = 29 , \mu F \times 225 , V)(Q2 = 6525 , \mu C)(Q2 = 6.525 \times 10^{-3} , C)2. Single Capacitor: C = 88µF, V = 530VWe can use the same formula (Q = CV) to find the charge stored in this capacitor.Charge (Q): (Q = C \times V = 88 , \mu F \times 530 , V) (Q = 46640 , \mu C) (Q = 4.6640 \times 10^{-2} , C)3. Series Capacitors: C1 = 36µF, C2 = 36µF, C3 = 36µF (in series)For capacitors in series, the equivalent capacitance ((C_{eq})) is given by:[\frac{1}{C_{eq}} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C3}]Equivalent Capacitance ((C_{eq})): [\frac{1}{C_{eq}} = \frac{1}{36 , \mu F} + \frac{1}{36 , \mu F} + \frac{1}{36 , \mu F} = \frac{3}{36 , \mu F} = \frac{1}{12 , \mu F}] [C_{eq} = 12 , \mu F]4. Series Circuit: C1 = 1.5F, C2 = 2.2F, C3 = 6.9F, C4 = 8.6F, V = 1000VIn a series capacitor circuit, the charge (Q) on each capacitor is the same and equal to the charge on the equivalent capacitance. First, we find the equivalent capacitance ((C_{eq})):[\frac{1}{C_{eq}} = \frac{1}{1.5 , F} + \frac{1}{2.2 , F} + \frac{1}{6.9 , F} + \frac{1}{8.6 , F}][\frac{1}{C_{eq}} \approx 0.6667 + 0.4545 + 0.1449 + 0.1163 = 1.3824 , F^{-1}][C_{eq} \approx \frac{1}{1.3824} , F \approx 0.7234 , F]Now, we can find the charge (Q) on the equivalent capacitance (and thus on each individual capacitor):[Q = C_{eq} \times V = 0.7234 , F \times 1000 , V = 723.4 , C]Next, we find the voltage across each capacitor using (V = \frac{Q}{C}):Voltage across C1 (V1): [V1 = \frac{Q}{C1} = \frac{723.4 , C}{1.5 , F} \approx 482.27 , V]Voltage across C2 (V2): [V2 = \frac{Q}{C2} = \frac{723.4 , C}{2.2 , F} \approx 328.82 , V]Voltage across C3 (V3): [V3 = \frac{Q}{C3} = \frac{723.4 , C}{6.9 , F} \approx 104.84 , V]Voltage across C4 (V4): [V4 = \frac{Q}{C4} = \frac{723.4 , C}{8.6 , F} \approx 84.12 , V]Charges: (Q1 = Q2 = Q3 = Q4 = 723.4 , C)Voltages: (V1 \approx 482.27 , V), (V2 \approx 328.82 , V), (V3 \approx 104.84 , V), (V4 \approx 84.12 , V)(Note that the sum of the voltages should approximately equal the battery voltage: (482.27 + 328.82 + 104.84 + 84.12 \approx 999.05 , V), the slight difference is due to rounding.)5. Total Resistance in Number 4 in 1 Day in Scientific Notation.Question number 4 describes a circuit with capacitors, not resistors. Therefore, the concept of resistance doesn't apply directly. If you meant to ask about something else related to that circuit over time, please clarify.However, if you are asking for the number of seconds in one day in scientific notation:1 day = 24 hours1 hour = 60 minutes1 minute = 60 secondsNumber of seconds in 1 day = (24 \times 60 \times 60 = 86400) secondsIn scientific notation: (8.64 \times 10^{4}) seconds.6. Capacitance needed to store 3.5µC at 256V.We can use the formula (Q = CV) and rearrange it to solve for capacitance ((C)):[C = \frac{Q}{V}]Capacitance (C): [C = \frac{3.5 , \mu C}{256 , V}] [C \approx 0.01367 , \mu F]In scientific notation: (1.367 \times 10^{-2} , \mu F) or (1.367 \times 10^{-8} , F).
