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Find the time constant of a capacitor with capacitance of 2 microfarad having an internal resistance of 4 megaohm.
- a)2
- b)0.5
- c)8
- d)0.25
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Find the time constant of a capacitor with capacitance of 2 microfarad...
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.
Most Upvoted Answer
Find the time constant of a capacitor with capacitance of 2 microfarad...
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.
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Community Answer
Find the time constant of a capacitor with capacitance of 2 microfarad...
To find the time constant of a capacitor with a given capacitance and internal resistance, we use the formula:
Time constant (τ) = Capacitance (C) * Internal resistance (R)
Given:
Capacitance (C) = 2 μF
Internal resistance (R) = 4 MΩ
Substituting these values into the formula, we get:
Time constant (τ) = 2 μF * 4 MΩ
Before multiplying, we need to convert the units to a common base. Since 1 F = 10^6 μF and 1 Ω = 10^6 MΩ, we can convert the units as follows:
2 μF = 2 * 10^-6 F
4 MΩ = 4 * 10^6 Ω
Substituting these converted values into the formula:
Time constant (τ) = (2 * 10^-6 F) * (4 * 10^6 Ω)
Multiplying the values together, we get:
Time constant (τ) = 8 * 10^0 FΩ
Simplifying further, we have:
Time constant (τ) = 8 FΩ
Now, we need to express the time constant in a more convenient unit. Since the farad (F) and ohm (Ω) are quite large units, we can use their prefixes to express the time constant in a more manageable form. The prefix "mega" (M) represents a factor of 10^6.
Converting the time constant to megaohm-farads:
Time constant (τ) = 8 * 10^-6 MΩF
Therefore, the time constant of the given capacitor is 8 μs or 8 * 10^-6 seconds. This corresponds to option (c) in the given choices.
Time constant (τ) = Capacitance (C) * Internal resistance (R)
Given:
Capacitance (C) = 2 μF
Internal resistance (R) = 4 MΩ
Substituting these values into the formula, we get:
Time constant (τ) = 2 μF * 4 MΩ
Before multiplying, we need to convert the units to a common base. Since 1 F = 10^6 μF and 1 Ω = 10^6 MΩ, we can convert the units as follows:
2 μF = 2 * 10^-6 F
4 MΩ = 4 * 10^6 Ω
Substituting these converted values into the formula:
Time constant (τ) = (2 * 10^-6 F) * (4 * 10^6 Ω)
Multiplying the values together, we get:
Time constant (τ) = 8 * 10^0 FΩ
Simplifying further, we have:
Time constant (τ) = 8 FΩ
Now, we need to express the time constant in a more convenient unit. Since the farad (F) and ohm (Ω) are quite large units, we can use their prefixes to express the time constant in a more manageable form. The prefix "mega" (M) represents a factor of 10^6.
Converting the time constant to megaohm-farads:
Time constant (τ) = 8 * 10^-6 MΩF
Therefore, the time constant of the given capacitor is 8 μs or 8 * 10^-6 seconds. This corresponds to option (c) in the given choices.
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