The equivalent capacitance (
) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.
<h3>The types of circuit.</h3>
Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;
<h3>What is a parallel circuit?</h3>
A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance () of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.
Mathematically, the equivalent capacitance () of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:
Ceq = C₁ + C₂ + C₃ + C₄
Substituting the given parameters into the formula, we have;
Ceq = 10 F + 3 F + 7 F + 1 F
Equivalent capacitance, Ceq = 21 F.
Read more equivalent capacitance here: brainly.com/question/27548736
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Answer:
n = 1.4
Explanation:
Given,
R1 = 18 cm, R2 = -18 cm
From lens makers formula
1/f = (n - 1)(1/18 + 1/18) = (n-1)/9
f = 9/(n-1)
Power, P = 1/f ( in m) = (n-1)/0.09
Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens
Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09
Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05
n = 1.4
