A capacitor of capacitance C1 = 1μF can with stand maximum voltage V1 = 6kV (kilo-volt) and another capacitor of capacitance C2 = 3μF can withstand maximum voltage V2 = 4 kV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

As Q = CV, (Q1)max=10–6 × 6 × 103 = 6mC While (Q2)max= 3 × 10–6 × 4 × 103 = 12mC However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 mC) and potential difference across it V2 = 6mC/3 μF = 2KV and as in series V =V1 + V2 So Vmax= 6KV + 2KV = 8KV

A capacitor of capacitance C1 = 1μF can with stand maximum voltage V1 = 6kV (kilo-volt) and another capacitor of capacitance C2 = 3μF can withstand maximum voltage V2 = 4 kV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

As Q = CV, (Q1)max=10–6 × 6 × 103 = 6mC While (Q2)max= 3 × 10–6 × 4 × 103 = 12mC However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 mC) and potential difference across it V2 = 6mC/3 μF = 2KV and as in series V =V1 + V2 So Vmax= 6KV + 2KV = 8KV

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A capacitor of capacitance C₁ = 1μF can with stand maximum voltage V₁ = 6kV (kilo-volt) and another capacitor of capacitance C₂ = 3μF can withstand maximum voltage V₂ = 4 kV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

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4 kV

6 kV

8 kV

10 kV

answer is C.

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Detailed Solution

As Q = CV, (Q₁)_max=10^–6 × 6 × 10³ = 6mC
While (Q₂)_max= 3 × 10^–6 × 4 × 10³ = 12mC

However in series charge is same so maximum charge on C₂ will also be 6 mC (and not 12 mC) and potential difference across it V₂ = 6mC/3 μF = 2KV and as in series V =V₁ + V₂ So V_max= 6KV + 2KV = 8KV

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