The peak current through a capacitor is 10.0 mA. What is the current if a. The emf frequency is doubled? b. The emf peak voltage is doubled (at the original frequency)? c. The frequency is halved and, at the same time, the emf is doubled?

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The peak current through a capacitor is 10.0 mA. What is the current if   a. The emf frequency is doubled?   b. The emf peak voltage is doubled (at the original frequency)?   c. The frequency is halved and, at the same time, the emf is doubled?

macalpinee3 · Accepted Answer

Step 1   The reactance of capacitor is;  XC=12πfC  Let potential is V volts.   So, current is given by;  l=VXC  =V12πfC  =2πfCV  Step 2   (a) Since `current is directly proportional to frequency.   So, when emf frequency is doubled then current is also doubled.   Hence, peak current will be 20.0 mA.   (b) Since, I=2πfCV this shows current is directly proportional to emf voltage.   So, when emf voltage is doubled then current also doubled.   Hence, peak current will be 20.0 mA.   (c) Since, I=2πfCV this shows current is directly proportional to emf voltage and frequency.   So, when frequency is halved and emf voltage is doubled  then net effect on current remains same.   Hence, peak current will be 10.0 mA.

lalilulelo2k3eq · Accepted Answer

The peak current in a inductor circuit is defined as:  Ipeak=ξpeakXL  Ipeak=ξpeak2⋅π⋅f⋅L=10mA  a) If we doubled the emf frequency, the current peak will be:  Ipeak=ξpeak2⋅π⋅2⋅f⋅L  Ipeak=5mA  If we double emf frequency, the peak current will be halved.  b) If we doubled the emf voltage, the current peak will be:  Ipeak=2⋅ξpeak2⋅π⋅f⋅L  Ipeak=20mA  If we double emf voltage, the peak current will be doubled.  c) If we doubled the emf voltage, and halved the emf frequency, the current peak will be:  Ipeak=2⋅ξpeak2⋅π⋅2⋅f⋅L  Ipeak=10mA  If we double emf voltage and half the emf frequency, the peak current will be the same.

alenahelenash · Accepted Answer

a. Explanation:Substitute 2f for f&#039; in equation (4),I′c=2π(2f)CVCI′c=2(2πfCVC)(5)Substitute equation (3) in (5),I′c=2IC (6)Substitute 10.0mA for IC in equation (6),I′c=2(10.0mA)=20.0mAIt can be concluded that if the new value of emf frequency (f&#039;) is doubled, then the new value of peak current throught the capacitor (I&#039;c) also doubled. Therefore, the new value is 20.0mA.Conclusion:Thus, the new value of peak current through the capacitor is 20.0mA.b. Explanation:Substitute 2VC for V&#039;c in equation(7),I′c=2πfC(2VC)I′c=2(2πfCVC)(8)Substitute equation (3) in (8),I′c=2IC (9)Substitute 10.0 mA for IC in equation (9),I′c=2(10.0mA)=20.0mAIt can be concluded that if the new value of emf peak voltage (V&#039;c) is doubled, then the new value of peak current through the capacitor (I&#039;c) also doubled. Therefore, the new value is 20.0 mA.Conclusion:Thus, the new value of peak current through the capacitor is 20.0mA.c. Explanation:Substitute 2VC for V&#039;c and f2 for f&#039; in equation (10),I′c=2π(f2)C(2VC)I′c=2πfCVC (11)Substitute equation (3) in (11),I′c=IC (12)Substitute 10.0mA for IC in equation (12),I′c=10.0mAIt can be concluded that if the new value of emf frequency (f&#039;) is halved and the new value of emf peak voltage (V&#039;c) is doubled, then the new value of peak current through the capacitor remains same.Conclusion:Thus, the new value of peak current through the capacitor is 10.0mA.

The peak current through a capacitor is 10.0 mA. What

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