voltage = resistance x current
= 5 x 2
= 10 volts
E = I R = (2) (6) = 12 volts
14v
Resistance = 11 ohmsCurrent = 2 amperesVoltage = 22 volts
a(r) = 2pi*r^2 and d = 2r 5 = 2r => r = 5/2 a(r) = 2pi*(5/2)^2 a(r) = 2pi*25/4 a(r) = 25pi/2 or, in decimal form, 39.2699 millimetres2
1) If it is: (r^22)(r^3) = r^(22 + 3) = r^25 (Just add powers) 2) If it is: (r^2)(2)(r^3) = 2[r^(2 + 3)] = 2r^5
E = I R = (2) (6) = 12 volts
14v
Can not do it without knowing the voltage I = E/R. Amps = Voltage/Ohms.
I = v/r = 10 / 5 = 2a.
If I0 = V/R, then Inew = (2*V)/(.5*R) = (2 / .5) * (V/R) = 4 *V/R = 4 * I0
That depends on what voltage it's designed to operate from. Power = (voltage)2 / R R = Voltage2 / power If it's a 117-volt bulb, R = (117)2 / 28 = 489 ohms. If it's a 240-volt bulb, R = (240)2 / 28 = 2,057 ohms.
ohms=amps/volts Amps= volts/ohms Volts = Amps*Ohms
Ohms law. R = E/I,where R= resistance in ohms, E = voltage in volts, and I = current in amperes.
The formula you are looking for is Ohm's Law. Voltage = Current x Resistance (v = I x R). To solve for Current the formula is I = V/R.
V=i*r=2*6=12v
V = I times R where V = voltage, I = current and R = resistance. Further, I = V / R.As I = V / R, I = 60 /12 = 5 amps.V=IR , where V=60 volts R=12 ohms so I = V/R = 60/12 = 5 Amp.
Resistance = 11 ohmsCurrent = 2 amperesVoltage = 22 volts