Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil
Answer:
6.169 μA
Explanation:
Formula for induced EMF is given by the equation;
EMF = M(di/dt). We are given;
di/dt = 2.5 A/s
M = 19μH = 19 × 10^(-6) H
Thus;
EMF = 19 × 10^(-6) × 2.5.
EMF = 47.5 × 10^(-6) V
Formula for current is;
i = EMF/R. R is resistance given as 7.7 ohms.
Thus; i = 47.5 × 10^(-6)/7.7
i = 6.169 μA
So we want to know what changes inside the multimeter when we change the voltage range from 200 V to 20 V, by what factor and does it increase or decrease. What we want when trying to measure the voltage with a multimeter is that a minimal current passes trough the mulitmeter so when we change the voltage range, we decrease the resistance by a factor of 10 because the voltage is decreased by a factor of 10.
Answer:
C. Quadruple
Explanation:
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I hope this helps! Have a great day!
The period of the wave would be halved
