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If a capacitor of capacitance 600 μF is charged at a uniform rate of 50 μC s⁻¹, what is the time required to increase its potential by 10 V?

Asked in GSEB Board July 2015 · Capacitance of a conductor and a capacitor

• (1) 500 s
• (2) 6000 s
• (3) 12 s
• (4) 120 s

Step-by-step solution

Why the other options are wrong

• (1) Multiplies the charging rate by the voltage, 50×10=500. That product has units of μC V s⁻¹, not seconds, and the capacitance never enters.
• (2) Stops at the charge needed, Q=C Δ V=600 μF×10 V=6000 μC, and reads it as a time. That charge still has to be divided by the rate of 50 μC per second.
• (3) Divides 600 by 50 and forgets the 10 V: 12 s is the time to raise the potential by just 1 V.

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