Question

a capacitor of 20 microfarads charged to 500v is connected in parallel with another capacitor of 10\(\mu F\) charged to 200V.Find the common potential.

Answer 1

When capacitors are connected in parallel, the potential difference across each capacitor is the same. Let's assume the common potential (voltage) across both capacitors is V. 

For the first capacitor, we have: Capacitance (C₁) = 20 μF Voltage (V1) = 500 V For the second capacitor, we have: Capacitance (C₂) = 10 μF Voltage (V₂) = 200 V Since the potential difference across both capacitors is the same (V), we can set up the following equations: Charge on the first capacitor (Q₁) = C₁\(\times\)V₁ Charge on the second capacitor (Q₂) = C₂\(\times\)V₂ Since the capacitors are connected in parallel, the total charge remains the same: Q₁ + Q₂ = Total Charge 

Substituting the values: C₁\(\times\)V₁ + C₂\(\times\) V₂ = Total Charge (20 μF) \(\times\) (500 V) + (10 μF) \(\times\) (200 V) = Total Charge 10000 μC + 2000 μC = Total Charge 12000 μC = Total Charge Now, the total charge is shared between the two capacitors. Since the capacitors are in parallel, the total charge is the sum of the charges on each capacitor: Total Charge = Q₁ + Q₂ 

Substituting the values: 12000 μC = C₁\(\times\)V + C₂ \(\times\) V 12000 μC = (20 μF) \(\times\)V + (10 μF) \(\times\)V 12000 μC = 30 μF \(\times\) V Now, solving for V (common potential): V = 12000 μC / 30 μF V = 400 V Hence, the common potential (voltage) across both capacitors is 400 V.