A 500 μ F capacitor is charged at the steady rate of 100 μ C/s . How long will it take to raise the potential difference between the plates of the capacitor to 10 V?

Question

A 500 μ F capacitor is charged at the steady rate of 100 μ C/s . How long will it take to raise the potential difference between the plates of the capacitor to 10 V? (A) 5s (B) 10s (C) 50s (D) 100s

Tardigrade Admin · Accepted Answer

Charge, Q=C V=500 μ F × 10 V =5 × 10-3 C Now, Q=q t or t=(Q/q) Or t=(5 × 10-3/100 × 10-6) s =(1/20) × 1000 s =50 s

Q. A $500 μ F$ capacitor is charged at the steady rate of $100 μ C / s$ . How long will it take to raise the potential difference between the plates of the capacitor to $10 V$ ?

5s

10s

50s

100s

Charge, $Q = C V = 500 μ F \times 10 V$
$= 5 \times 1 0^{- 3} C$
Now, $Q = qt$
or $t = \frac{Q}{q}$
Or $t = \frac{5 \times 1 0 ^{- 3}}{100 \times 1 0 ^{- 6}} s$
$= \frac{1}{20} \times 1000 s$
$= 50 s$

Q. A 500μF capacitor is charged at the steady rate of 100μC/s . How long will it take to raise the potential difference between the plates of the capacitor to 10V?

5s

10s

50s

100s

Charge, Q=CV=500μF×10V =5×10−3C Now, Q=qt or t=qQ​ Or t=100×10−65×10−3​s =201​×1000s =50s

Q. A $500 μ F$ capacitor is charged at the steady rate of $100 μ C / s$ . How long will it take to raise the potential difference between the plates of the capacitor to $10 V$ ?

Charge, $Q = C V = 500 μ F \times 10 V$
$= 5 \times 1 0^{- 3} C$
Now, $Q = qt$
or $t = \frac{Q}{q}$
Or $t = \frac{5 \times 1 0 ^{- 3}}{100 \times 1 0 ^{- 6}} s$
$= \frac{1}{20} \times 1000 s$
$= 50 s$