How many 1 μ F capacitors must be connected in parallel to store a charge of 1 C with a potential of 110 V across the capacitors?

Question

How many 1 μ F capacitors must be connected in parallel to store a charge of 1 C with a potential of 110 V across the capacitors? (A) 990 (B) 900 (C) 9090 (D) 909

Tardigrade Admin · Accepted Answer

charge q=n C V 1=n × 1 × 10-6 × 110 n =(1/110 × 10-6) n=(100000/11) ≈ 9090

Q. How many $1 μ F$ capacitors must be connected in parallel to store a charge of $1 C$ with a potential of $110 V$ across the capacitors?

990

900

9090

909

charge $q = n C V$
$1 = n \times 1 \times 1 0^{- 6} \times 110$
$n = \frac{1}{110 \times 1 0 ^{- 6}}$
$n = \frac{100000}{11}$
$\approx 9090$

Q. How many 1μF capacitors must be connected in parallel to store a charge of 1C with a potential of 110V across the capacitors?

990

900

9090

909

charge q=nCV 1=n×1×10−6×110 n=110×10−61​ n=11100000​ ≈9090

Q. How many $1 μ F$ capacitors must be connected in parallel to store a charge of $1 C$ with a potential of $110 V$ across the capacitors?

charge $q = n C V$
$1 = n \times 1 \times 1 0^{- 6} \times 110$
$n = \frac{1}{110 \times 1 0 ^{- 6}}$
$n = \frac{100000}{11}$
$\approx 9090$