Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg
1 answer:
Each capacitor carry the same charge 'q'.
Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:
= V₁ +V₂ +V₃
This voltage may also be calculated using capacitance and charge;
V = Q/ C
= V₁ +V₂ +V₃
Provided that the total charge is 'q', hence the total voltage can be expressed as:
= (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)
Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.
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Answer: Option (d) is correct.
Explanation:
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1 British thermal unit = 1055.06 joules
So, in 1,152 British thermal units there will be :
Hence, from the given options the closest answer is of option (d). So, option (d) is correct.
Answer:
ΔR = 9 s
Explanation:
To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value
The given expression is R = 2A / B
the uncertainty is ΔR = | | ΔA + |
| ΔB
we look for the derivatives
= 9 / B
= 9A (
)
we substitute
ΔR = ΔA +
ΔB
the values are
ΔA = 2 s
ΔB = 3 s
ΔR = 2 +
3
ΔR = 1.636 + 7.14
ΔR = 8,776 s
the absolute error must be given with a significant figure
ΔR = 9 s