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When a capacitor was connected across the terminals of an ohm meter, the pointer indicated a low resistance initially and then slowly came back to very high resistance. This indicates that capacitor is
- a)Faulty
- b)All right
- c)Leaky
- d)Short-circuited
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
When a capacitor was connected across the terminals of an ohm meter, t...
Explanation:
1. Initial Reading: When the capacitor is connected across the terminals of an ohm meter, the pointer indicates a low resistance initially. This is because the ohm meter initially charges the capacitor, causing the pointer to show a low resistance.
2. Slowly Comes Back to High Resistance: As the capacitor gets fully charged, the current flowing through it decreases, causing the pointer of the ohm meter to slowly come back to very high resistance. This is because a fully charged capacitor acts like an open circuit to direct current.
3. Indication of All Right: The slow transition of the pointer from low resistance to high resistance indicates that the capacitor is functioning correctly. This behavior is expected when testing a capacitor with an ohm meter.
4. Conclusion: Therefore, the ohm meter indicating a low resistance initially and then slowly coming back to very high resistance when a capacitor is connected across its terminals is an indication that the capacitor is all right and functioning as expected.
1. Initial Reading: When the capacitor is connected across the terminals of an ohm meter, the pointer indicates a low resistance initially. This is because the ohm meter initially charges the capacitor, causing the pointer to show a low resistance.
2. Slowly Comes Back to High Resistance: As the capacitor gets fully charged, the current flowing through it decreases, causing the pointer of the ohm meter to slowly come back to very high resistance. This is because a fully charged capacitor acts like an open circuit to direct current.
3. Indication of All Right: The slow transition of the pointer from low resistance to high resistance indicates that the capacitor is functioning correctly. This behavior is expected when testing a capacitor with an ohm meter.
4. Conclusion: Therefore, the ohm meter indicating a low resistance initially and then slowly coming back to very high resistance when a capacitor is connected across its terminals is an indication that the capacitor is all right and functioning as expected.
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Community Answer
When a capacitor was connected across the terminals of an ohm meter, t...
Concept:
Capacitor behavior:
- Ohmmeter uses dc voltage (battery) to measure the unknown resistance.
- Then the capacitor is subjected to dc source, initially, it acts as a short circuit and at steady-state, it acts as an open circuit.
- At the time when the ohmmeter is connected to the capacitor, the capacitor acts as a short circuit, so low resistance is indicated.
- After some time, the capacitor behaves as an open-circuit, so the pointer moves to the final position.
- This states that the capacitor is in very good condition to be used.
Important Points
- Capacitive reactance is given by Xc = 1/(2πfc) where f = frequency and c = value of capacitance
- Frequency f = 1/T (Time period)
- At the instant when the capacitor is connected to dc source, then V volt is reached in 0 sec so the time period is 0 sec.
- This makes the frequency to be infinity, therefore capacitive reactance becomes zero.
- That's why the capacitor acts as a short circuit initially when the dc source is applied.
- Now capacitor starts charging and voltage is developed across it.
- When the steady-state voltage across the capacitor is equal to source voltage, then after that it doesn't allow any current to pass through it.
- Therefore, acts as highly resistive or this can also be seen when at steady-state, frequency is zero, so capacitive reactance becomes infinity.
- That's why the capacitor blocks dc at the steady-state condition.
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When a capacitor was connected across the terminals of an ohm meter, the pointer indicated a low resistance initially and then slowly came back to very high resistance. This indicates that capacitor isa)Faultyb)All rightc)Leakyd)Short-circuitedCorrect answer is option 'B'. Can you explain this answer?
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