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Q = C V

C = Q / V = (60/12) = 5 farads (a capacitor the size of a house)

Q: What is the capacitance when Q equals 60 C and V equals 12 V?

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a = 20 b = 60 c = 100

If 8c = 60, then c = 60/8 = 7.5.

If we're talking strictly algebra. 12 equals p of c can be written as: 12= p(c) meaning, 12 is the answer for some function p, when c is the variable.

yes. b/c 12*5=60

t = 4/5

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C = Q/V; capacitance = charge / voltage. Therefore, you would also have to know the voltage.

C = capacitance, f = frequency ===> Capacitive reactance = 1 / [ 2(pi)fC ] 663 = 1 / [ 2(pi)(60)C ] 663 x 2 x pi x 60 x C = 1 C = 1 / (663 x 2 x pi x 60) = 1 / (663 x 120 x pi) = 1 / 249,945.1 = 4 x 10-6 = 4 microfarads (almost exactly)

a = 20 b = 60 c = 100

The value of c/30 that equals 12 is 12 . In order to make that value, 'c' must be 360 .

If 8c = 60, then c = 60/8 = 7.5.

If we're talking strictly algebra. 12 equals p of c can be written as: 12= p(c) meaning, 12 is the answer for some function p, when c is the variable.

yes. b/c 12*5=60

t = 4/5

C

-12 + c = 19 Add 12 to both sides: c = 19+12 = 31

space charge region in a diode or say a bjt for better understanding is same as the depletion region, both transition capacitance and depletion capacitance are the same c= (epsilon*A)/d ; where ... c is capacitance A is area and d is the depletion width the other type of capacitance is the diffusion capacitance c= (T*I)/(n*V) where ... c is the capacitance T is transition ti me I is the drift current n is emission coefficient ... its value is 1 for germanium and V is thermal voltage .. 26mv

12 deg F = -11.1 deg C